'■^'■J®*!-"- 


IN  MEMORIAM 
FLORIAN  CAJORl 


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A 


COURSE 

OF 

MATHEMATICS. 

IN  TWO  VOLUMES. 

FOR  THE  USE  OF  ACADEMIES, 

AS    WELL    AS 

PRIVATE  TUITION. 


BY 

CHARLES  HUTTON,  L.L.D.  F.R.a 

CATE    PROFESSOR    OF    MATHEMATICS     IHJ    THE    JROltAIiv    MILITARY 
ACADEMY. 


' '-'  'KZViblJD  AND  CdKRiSCTEO  l/T  ' 

ROBERT  ADRAIN,  A.  M. 

FELLOW  OF  THE  AMERICAN  PHILOSOPHICAL  SOCIETr, 

AND 

PROFESSOR  OF  MATHEMATICS  IN  QUEEN'S  COLLEGE  NEW-SERSEt. 


VOL.  I. 


NEW- YORK : 

PUBLISHED  BY  SAMUEL  CAMPBELL,  EVERT  DUYCKINCl^ 

T.  &  J.  SWORDS,  PETER  A.  MESIER,  R.  M'DERMUT, 

THOMAS  A.  RONALDS,  JOHN  TIEBOUT, 

AND  GEORGE  LONG. 

1818. 


ADVERTISEMEiNT. 


THE  publishers  of  this  third  AmericanjedUion  of  Dr.  Hutton'.s  Course  of 
Mathfc.natics,  were  ifTdncdTTo  engageT^n't'Ee^work  frora  a  conviction  of 
its  utility  to  private  Students,  as  well  as  to  Colleges  and  other  Seminaries,  in 
vhich  Mathematical  Science  constitutes  a  branch  of  education.  They  also 
had  in  view  the  furnishing  of  the  Military  School  of  their  country  with  a  Text 
Book  of  high  standing,  and  long  in  use  in  the  British  Military  Academy  And 
in  order  that  this  edition  might  derive  advantage  from  the  progress  of  the 
Science,  and  thereby  become  more  worthy  of  the  public  patronage,  they  en- 
gaged a  gentleman  of  acknowledged  eminence  to  revise  its  pages  and  super- 
intend the  printing  ;  and  they  confidently  trust  this  duty  has  been  performed 
•\viih  some  profit  to  the  work  generally.  To  gentlemen,  therefore,  who  study 
this  delightful  science  in  private,  and  to  the  literary  and  military  institutions  of 
their  country,  the  publishers  and  proprietors  look  for  remuneration — and  they 
feel  as  though  they  should  not  look  in  vain.  An  increasing  taste  for  Mathe- 
matical Studies  will  produce  a  correspondent  increase  of  purchasers  ;  whife 
the  preference  which  an  honourable  patriotism  gives  to  American  editions 
*when  well  executed,  will  receive  additional  activity  from  the  super-eminence 
tof  the  work  itself. 

Qfj=*  Oiders  for  this  publication  will  be  thankfully  received  by  any  of  the 
proprietors;  all  whose  names  are  printed  at  the  foot  of  the  title-page. 

jSTew-Yorlc,  1818.  X*  *    ' 


DUtrict.  nf  iJeivTork,  ss-  \ 

BE  IT  REMEMBERED,  that  on  the  eleventh  day  of  August,  in  the  tliiity-seventli  year 
of  the  Independence  of  the  United  States  of  America,  Samuel  Campbell^  of  the  said  district, 
liath  deposited  m  this  office  tlie  title  of  a  book,  the  right  whereof  he  claims  as  proprietor, 
hi  the  woi-ds  following:,  to  wit: 

"  A  Com'se  of  Mathematics.  In  two  vgjuipjes.  For  the  use  of  Academies,  as  well  as  Pri- 
vate Tuition.  By  Charles  HiUtoiiI  L.  L.  D.  F.  R.  S.  late  Professor  of  Mathematics  in  the 
Royal  Military  Academy.  From  the  fifth  and  sixth  London  editions,  revised  and  correctefl 
by  Robert  Atlrain,  A.  M.  Fellow  of  the  American  Philosophical  Society,  and  Professor  of 
Mathematics  in  Queen's  College.  New-Jersey." 

In  conformity  to  the  Act  of  the  Congress  of  the  United  States,  entitled  "  An  act  for  the 
encouvaRement  of  learning,  by  securing  the  copies  of  maps,  charts,  and  books  to  the' au- 
thors and  iiroprietors  ol'  such  copies^  during  the  times  therein  mentioned."  And  also  to  an 
act,  entitletl  "  An  act<  supplementary  to  an  act,  entitled  an  act  for  the  encouragement  of 
learning,  by  securing  the  copies  of  maps,  charts,  and  books  to  the  authors  and  proprietors 
of  such  coi'ies,  during  the  times  therein  mentioned,  and  extending  the  heaefits  thereof  to 
<he  arts  of  designing,  engva-y  ing,  and  etching  historical  and  other  prints." 

CHARLES  CLINTON, 
Ctcrh  of  the  district  of  NewTarE;. 


Gem^ge  Long,  Printer. 


CAJOR? 


PREFACE. 


A  SHORT  and  Easy  Course  of  the  Mathematical  Sciences 
has  long  been  considered  as  a  desideratum  for  the  use  of  Stu- 
dents in  the  different  schools  of  education  :  one  that  should 
hold  a  middle  rank  between  the  more  voluminous  and  bulky 
collections  of  this  kind,  and  the  mere  abstract  and  brief  com- 
mon-place forms,  of  principles  and  memorandums. 

For  long  experience,  in  all  Seminaries  of  Learning,  has 
shown,  that  such  a  work  was  very  much  wanted,  and  would 
prove  a  great  and  general  benefit ;  as,  for  want  of  it,  recourse 
has  always  been  obliged  to  be  had  to  a  number  of  other  books 
hy  different  authors  ;  selecting  a  part  from  one  and  a  part  from 
another,  as  seemed  most  suitable  to  the  purpose  in  hand,  and 
rejecting  the  other  parts — a  practice  which  occasioned  much 
expense  and  trouble,  in  procuring  and  using  such  a  number 
of  odd  volumes,  of  various  forms  and  modes  of  composition  ; 
besides  wanting  the  benefit  of  uniformity  and  reference,  which 
are  found  in  a  regular  series  of  composition. 

To  remove  these  inconveniences,  the  Author  of  the  present 
work  has  been  induced,  from  time  to  time,  to  compose  various 
parts  of  this  Course  of  Mathematics  ;  which  the  experience 
of  many  years'  use  in  the  Academy  has  enabled  him  to  adapt 
and  improve  to  the  most  useful  form  and  quantity,  for  the  be- 
nefit of  instruction  there.  And,  to  render  that  benefit  mores 
eminent  and  lasting,  the  Master  General  of  the  Ordnance  has 
been  pleased  to  give  it  its  present  form,  by  ordering  it  to  bq 
enlarged  and  printed,  for  the  use  of  the  Royal  Military  Aca- 
demy. 

As  this  work  has  been  composed  expressly  with  the  inten- 
tion of  adaptmg  it  to  the  purposes  of  academical  education,  it 
is  not  designed  to  hold  out  the  expectation  of  an  entire  new 
mass  of  inventions  and  discoveries  :  but  rather  to  collect  and 
.  arrange  the  most  useful  known  principles  of  mathematics,  dis- 
posed in  a  convenient  practical  form,  demonstrated  in  a  plain 
and  concise  way,  and  illustrated  with  suitable  examples,  re- 
jecting whatever  seemed  to  be  matters  of  mere  curiosity,  and 
retaining  only  such  parts  and  branches,  as  have  a  direct  tenden- 
cy and  application  to  some  useful  purpose  in  life  or  profession. 
It  is  however  expected  that  much  that  is  new  will  be  found 
in  many  parts  of  these  volumes  ;  as  well  in  the  matter,  as  in 
the  arrangement  and  manner  of  demonstration,  throughout  the 
whole  work,  especially  in  the  geometry,  which  is  rendered 
much  more  easy  and  simple  thaa  heretgfore  ;  and  in  the  conic 

918242  ""'""' 


Vv  PREFACE. 

sections,  which  are  here  treated  in  a  manner  at  once  new,  easy 
and  natural ;  so  much  so  indeed,  that  all  the  propositions  and 
their  demonstrations,  in  the  ellipsis,  are  the  very  same,  word 
for  word,  as  those  in  the  hyperbola,  using  only,  in  a  very  few 
places,  the  word  swm,  for  the  word  difference  :  also  in  many  of 
the  mechanical  and  philosophical  parts  which  follow,  in  the 
second  volume.  In  the  conic  sections,  too,  it  may  bejobserved, 
that  the  first  theorum  of  each  section*only  is  proved  from  the 
cone  itself,  and  all  the  rest  of  the  theorems  are  deduced  from 
the  first,  or  from  each  other,  in  a  very  plain  and  simple  manner. 
Besides  renewing  most  of  the  rules,  and  introducing  every 
where  new  examples,  this  edition  is  much  enlarged  in  several 
places  ;  particularly  by  extending  the  tables  of  squares  and 
cubes,  square  roots  and  cube  roots,  to  1000  numbers,  which 
will  be  found  of  great  use  in  many  calculations  ;  also  by  the 
tables  of  logarithms,  sines,  and  tangents,  at  the  end  of  the  se- 
cond volume  ;  by  the  addition  of  Cardan's  rules  for  resolving 
'cubic  equations  ;  with  tables  and  rules  for  annuities  ;  and  many 
other  improvements  in  different  parts  of  the  work. 

Though  the  several  parts  of  this  course  of  mathematics  are 
ranged  in  the  order  naturally  required  by  such  elements,  yet 
students  may  omit  any  of  the  particulars  that  may  be  thought 
the  least  necessary  to  their  several  purposes  ;  or  they  may, 
study  and  learn  various  parts  in  a  different  order  from  their 
jpresent  arrangement  in  the  book,  at  the  discretion  of  the  tutor. 
So,  for  instance,  all  the  notes  at  the  foot  of  the  pages  may  be 
omitted,  as  well  as  many  of  the  rules  ;  particularly  the  1st  or 
Common  Rule  for  the  Cube  Root,  p.  85,  may  wel\  be  omitted, 
I)eing  more  tedious  than  useful.  Also  the  chapters  on  Surds 
and  Infinite  Series,  in  the  Algebra  :  or  these  might  be  learned 
after  Simple  Equations.  Also  Compound  Interest  and  Annui- 
ties at  the  end  of  the  Algebra.  Also  any  part  of  the  Geome- 
try, in  vol.  1  ;  any  of  the  branches  in  vol.  2,  at  the  discretion  of 
the  preceptor.  And,  in  any  of  the  parts,  he  may  omit  some  of 
the  examples,  or  he  may  give  more  than  are  printed  in  the 
book  ;  or  he  may  very  profitably  vary  or  change  them,  by  alter- 
ing the  numbers  occasionally. — As  ta  the  quantity  of  writing  •,. 
the  author  would  recommend,  that  the  student  copy  out  into 
his  fair  book  no  more  than  the  chief  rules  which  he  is  directed 
to  lejarn  off  by  rote,  with  the  work  of  one  example  only  to. 
each  rule,  set  down  at  full  length  ;  omitting  to  set  down  the 
work  of  all  the  other  examples,  how  many  soever  he  may  be 
directed  to  work  out  upon  his  slate  or  waste  paper. — In  short, 
a  great  deal  of  the  business,  as  to  the  quantity  and  order  and 
^  manner,  must  depend  on  the  judgment  of  the  discreet  and  pru- 
dent tutor  or  director. 

[Dr. 


[Dr.  Hutton's  Frefaee  to  the  third  voiurae  of  the  BoglijOi  edition, 
published  in  1811.] 


1  HE  beneficial  improvements  lately  made,  and  still  makiog^^ 
in  the  plan  of  the  scientific  education  of  the  Cadets,  in  the 
Royal  Military  Academy  at  Woolwich,  having  rendered  a 
further  extension  of  the  Mathematical  Course  adviseable,  I 
was  honoured  with  the  orders  of  his  Lordship  the  Master 
General  of  the  Ordnance,  to  prepare  a  third  volume,  in  addi- 
tion  to  the  two  former  volumes  of  the  Course,  to  contain 
such  additions  to  some  of  the  subjects  before  treated  of  in 
those  two  volumes,  with  such  other  new  branches  of  military 
science,  as  might  appear  best  adapted  to  promote  the  ends  of 
this  important  institution.  From  my  advanced  age,  and  the 
precarious  state  of  my  health,  I  was  desirous  of  declining  such 
a  task,  and  pleaded  my  doubts  of  being  able,  in  such  a  state, 
to  answer  satisfactorily  his  lordship's  wishes.  This  difficulty 
however  was  obviated  by  the  reply,  that,  to  preserve  a  uni- 
formity between  the  former  and  the  additional  parts  of  the 
Course,  it  was  requisite  that  I  should  undertake  the  direction 
of  the  arrangement,  and  compose  such  parts  of  the  work  as 
might  be  found  convenient,  or  as  related  to  topics  in  which 
I  had  made  experiments  or  improvements  ;  and  for  the  rest, 
I  might  take  to  my  assistance  the  aid  of  any  other  person  I 
might  think  proper.  With  this  kind  indulgence  being  en- 
couraged to  exert  my  best  endeavours,  1  immediately  an- 
nounced my  wish  to  request  the  assistance  of  Dr.  Gregory  of 
the  Royal  Military  Academy,  than  whom,  both  for  his  ex- 
tensive scientific  knowledge,  and  his  long  experience,  I  know 
«f  no  person  more  fit  to  be  associated  in  the  due  performance 
©f  such  a  task.  Accordingly,  this  volume  is  to  be  considered 
as  the  joint  composition  of  that  gentleman  and  myself,  hay- 
ing each  of  us  taken  and  prepared,  in  nearly  equal  portions^ 
separate  chapters  and  branches  of  the  work,  being  such  as 
in  the  compass  of  this  volume,  with  the  advice  and  assistance 
of  the  Lieut.  Governor,  were  deemed  among  the  most  useful 
additional  subjects  for  the  purposes  of  the  education  estab- 
lished in  the  Academy. 

The  several  parts  of  the  work,  and  their  arrangement,  are 
as  follow. — In  the  first  chapter  are  contained  all  the  proposi- 
tions of  the  course  of  Conic  Sections,  first  printed  for  the  use 
of  the  Academy  in  the  year  1767,  which  remained,  after 
those  that  were  selected  for  the  second  volume  of  this  Course  : 

to 


yi  PREFACE. 

to  which  is  added  a  tract  on  the  algebraic  equations  of  the 
several  conic  sections,  serving  as  a  brief  introduction  to  the 
algebraic  properties  of  curve  lines. 

The  2d  chapter  contains  a  short  geometrical  treatise  on  the 
elements  of  Isoperimetry  and  the  maxima  and  minima  of 
surfaces  and  solids ;  in  which  several  propositions  usually  in- 
vestigated by  fluxionary  processes  are  effected  geometrically  ; 
and  in  which,  indeed,  the  principal  results  deduced  by  Thos. 
Simpson,  Horsley,  Legendre,  and  Lhuillier  are  thrown  into 
the  compass  of  one  short  tract. 

The  3d  and  4th  chapters  exhibit  a  concise  but  compre- 
hensive view  of  the  trigonometrical  analysis,  or  that  in  which 
the  chief  theorems  of  Plane  and  Spherical  Trigonometry  are 
deduced  algebraically  by  means  of  what  is  commonly  de- 
nominated the  Arithmetic  of  Sines.  A  comparison  of  the 
modes  of  investigation  adopted  in  these  chapters,  and  those 
pursued  in  that  part  of  the  second  volume  ■  of  this  course 
which  is  devoted  to  Trigonometry,  will  enable  a  student  to 
trace  the  relative  advantages  of  the  algebraical  and  geome- 
trical methods  of  treating  this  useful  branch  of  science.  The 
fourth  chapter  includes  also  a  disquisition  on  the  nature  and 
measure  of  solid  angles,  in  which  the  theory  of  that  peculiar 
class  of  geometrical  magnitudes  is  so  represented,  as  to  ren- 
der their  mutual  comparison  (a  thing  hitherto  supposed  im- 
possible except  in  one  or  two  Very  obvious  cases)  a  matter  of 
perfect  ease  and  simplicity. 

Chapter  the  fifth  relates  to  Geodesic  Operations,  and  that 
more  extensive  kind  of  Trigonometrical  Surveying  which  is 
employed  with  a  view  to  determine  the  geographical  situation 
«f  places,  the  magnitude  of  kingdoms,  and  the  figure  of  the 
earth.  This  chapter  is  divided  into  two  sections  ;  in  the  first 
of  which  is  presented  a  general  account  of  this  kind  of  survey- 
ing ;  and  in  the  second,  solutions  of  the  most  important  prob- 
lems connected  with  these  operations.  ,  This  portion  of  the 
volume  it  is  hoped  will  be  found  highly  useful ;  as  there  is  no 
work  which  contains  a  concise  and  connected  account  of  this 
kind  of  surveying  and  its  dependent  problems  ;  and  it  cannot 
fail  to  be  interesting  to  those  who  know  how  much  honour  re- 
dounds to  this  country  from  the  great  skill,  accuracy,  and 
judgment,  with  which  the  trigonometrical  survey  of  England 
has  long  been  carried  on. 

In  the  6th  and  7th  chapters  are  developed  the  principles  of 
Pblygonometry.)  and  those  which  relate  to  the  Dimsion  of  lands 
and  other  surfaces,  both  by  geoaaetrical  coastructiou  and  by 
coDM^utatioa. 


PREFACE.  ^  tii 

The  8th  chapter  contains  a  view  of  the  nature  and  solution 
of  equations  in  general,  with  a  selection  of  the  best  rules  for 
equations  of  different  degrees.  Chapter  the  9th  is  devoted  to 
the  nature  and  properties  of  curves^  and  the  construction  of 
equations.  These  chapters  are  manifestly  connected,  and 
show  how  the  mutual  relations  subsisting  between  equations  of 
different  degrees,  and  curves  of  various  orders,  serve  for  the 
reciprocal  illustration  of  the  properties  of  both. 

In  the  1 0th  chapter  the  subjects  of  Fluents  and  Fluxional 
equations  are  concisely  treated.  The  various  forms  of  Fluents 
comprised  in  the  useful  table  of  them  in  the  2d  volume,  are 
investigated  :  and  several  other  rules  are  given  ;  such  as  it  is 
believed  will  tend  much  to  facilitate  the  progress  of  students 
in  this  interesting  department  of  science,  especially  those 
which  relate  to  the  mode  of  finding  fluents  by  continuation. 

The  11th  chapter  contains  solutions  of  the  most  useful 
problems  concerning  the  maximum  ejects  of  machines  in  mo- 
tion;  and  developes  those  principles  which  should  constantly 
be  kept  in  view  by  those  who  would  labour  beneficially  for  the 
improvement  of  machines. 

In  the  1 2th  chapter  will  be  found  the  theory  of  the  pres- 
sure of  earth  and  fluids  against  walls  and  fortifications  ;  and 
the  theory  which  leads  to  the  best  construction  of  powder 
magazines  with  equilibrated  roofs. 

The  13th  chapter  is  devoted  to  that  highly  interesting  sub- 
ject, as  well  to  the  philosopher  as  to  mihtary  men,  the  theory 
and  practice  of  gunnery.  Many  of  the  difficulties  attending 
this  abstruse  enquiry  are  surmounted  by  assuming  the  results 
of  accurate  experiments,  as  to  the  resistance  experienced  by 
bodies  moving  through  the  air,  as  the  basis  of  the  computa- 
tions. Several  of  the  most  useful  problems  are  solved  by 
means  of  this  expedient,  with  a  facility  scarcely  to  be  expect- 
ed, and  with  an  accuracy  far  beyond  our  most  sanguine  ex- 
pectations. 

The  14th  and  last  chapter  contains  a  promiscuous  but  ex- 
tensive collection  of  problems  in  statics,  dynamics,  hydrosta- 
tics, hydraulics,  projectiles,  &c.  &c.  ;  serving  at  once  to  exer- 
cise the  pupil  in  the  various  branches  of  mathematics  com- 
prised in  the  course,  to  demonstrate  their  utility  especially 
to  those  devoted  to  the  military  profession,  to  excite  a  thirst 
for  knowledge,  and  in  several  important  respects  to  gratify  it. 

This  volume. being  professedly  supplementary  to  the  pre- 
ceding two  volumes  of  the  Course  may  best  be  used  in  tuition 
by  a  kind  of  mutual  incorporation  of  its  contents  with  those 
of  the  second  volume.  The  method  of  effecting  this  will,  of 
course,  vary  according  to  eirctunstances,  and  the  precise  em- 
ployments 


viii  PREFACE. 

ployments  for  which  the  pupils  are  destined  :  but  in  general 
it  is  presumed  the  following  may  be  advantageously  adopted. 
Let  the  first  seven  chapters  be  taught  immediately  after  the 
Conic  Sections  in  the  2d  volume.     Then  let  the  substance  of 

^  the  2d  volume  succeed,  as  far  as  the  Practical  Exercises  on 
Natural  Philosophy,  inclusive.  Let  the  8th  and  9th  chapters 
in  this  3d  vol.  precede  the  treatise  on  Fluxions  in  the  2d  ;  and 
when  the  pupil  has  been  taught  the  part  relating  to  fluents  in 
that  treatise,  let  him  immediately  "he  conducted  through  the 
10th  chapter  of  the  3d  volume.  After  he  has  gone  over  the 
remainder  of  the  Fluxions  with  the  applications  to  tangents, 
Fadii  of  curvature,  rectifications,  quadratures,  &c.  the  1 1th  and 
12th  chapters  of  the  3d  vol.  should  be  taught.  The  prob- 
lems in  the  13th'  and  14th  chapters  must  be  blended  with  the 
practical  exercises  at  the  end  of  the  2d  volume,  in  such  manner 
as  shall  be  found  best  suited  to  the  capacity  of  the  student,  and 
best  calculated  to  ensure  his  thorough  comprehension  of  the 
several  curious  problems  contained  in  those  portions  of  the  work. 
In  the  composition  of  this  3d  volume,  as  weH  as  in  that  of 
the  preceding  parts  of  the  Course,  the  great  object  kept  con- 
stantly in  view  has  been  utility,  especially  to  gentlemen  in- 
tended for  the  Military  Profession.  To  this  end,  all  such  in- 
vestigations as  might  serve  merely  to  display  ingenuity  or  tal- 
ent, without  any  regard  to  practical  benefit,  have  been  carefully 
excluded.  The  student  has  put  into  his  hands  the  two  power- 
ful instruments  of  the  ancient  and  the  modern  or  sublime  ge- 
ometry ;  he  is  taught  the  use  of  both,  and  their  relative  ad- 
vantages are  so  exhibited  as  to  guard  him,  it  is  hoped,  from 
any  undue  and  exclusive  preference  for  either.  Much  novel- 
ty of  matter  is  not  to  be  expected  in  a  work  like  this  ;  though, 
considering  its  magnitude,  and  the  frequency  with  which  sev- 
eral of  the  subjects  have  been  discussed,  a  candid  reader  will 

'  not,  perhaps,  be  entirely  disappointed  in  this  respect.  Per- 
spicuity and  condensation  have  been  uniformly  aimed  at  through 
the  performance  :  and  a  small  clear  type,  with  a  full  page,  have 
been  chosen  for  the  introduction  of  a  large  quantity  of  matter. 
A  candid  public  will  accept  as  an  apology  for  any  slight  dis- 
order or  irregularity  that  may  appear  in  the  composition  and 
arrangement  of  this  Course,  the  circumstance  of  the  different 
volumes  having  been  prepared  at  widely  distant  times,  and 
with  gradually  expanding  views.  But,  on  the  whole,  I  trust  it 
will  be  found  that,  with  the  assistance  of  my  friend  and  coad- 
jutor in  this  supplementary  volume,  I  have  now  produced  a 
Course  of  Mathematics,  in  which  a  great  variety  of  useful  sub- 
jects are  introduced,  and  treated  with  perspicuity  and  correct-' 
ness,  thanin  any  three  volumes  of  equal  size  in  any  language. 

CHA.  HUTXOK. 


PREFACE, 

BY  THE  AMERICAN  EDITOR. 


1  HE  last  English  edition  of  HuttorCs  Course  of  Mathema- 
tics, in  three  volumes  octavo,  may  be  considered  as  one  of  the 
best  systems  of  Mathematics  in  the  English  language.  Its 
great  excellence  consists  in  the  judicious  selection  made  by 
the  authors  of  the  work,  who  have  constantly  aimed  at  such 
things  as  are  most  necessary  in  the  useful  arts  of  life.  To 
this  may  be  added  the  easy  and  perspicuous  manner  in  which 
the  subject  is  treated — a  quality  of  primary  importance  in  a 
treatise  intended  for  beginners,  and  containing  the  elements 
of  science. 

TheJh[r4js^luaie^^X--lhB^  EBgli»h..€^^^ 
lately  published,  is  scarcely  known  at  present  in  this  country- 
ins  but  justice  Itb  its  excellent  authors  to  state,  that  they  have 
collected  in  it  a  great  number  of  the  most  interesting  sub- 
jects in  Analytical  and  Mechanical  vScience.  Analytical  Tri- 
gonometry Plane  and  Spherical,  Trigonometrical  Surveying, 
Maxima  and  Minima  of  Geometrical  Quantities,  Motion  of 
Machines  and  their  Maximum  Effects,  Practical  Gunnery, 
&c.  are  among  the  most  important  subjects  in  Mathematics, 
and  are  disciissed  in  the  volume  just  mentioned  in  such  a 
manner  as  not  only  to  prove  highly  useful  to  pupils,  but  also 
to  such  as  are  engaged  in  various  departments  of  Practical 
Science. 

As  the  work,  after  the  publication  of  the  third  volume,  em- 
braced most  subjects  of  curiosity  or  utihty  in  Mathematics, 
it  has  been  thought  unnecessary  to  enlarge  its  size  by  much 
additional  master.  The  present  edition  however,  differs  in 
several  respects  from  the  last  English  one  ;  and  it  is  presum- 
ed, that  this  difference  will  be  found  to  consist  of  improve- 
ments.    These  are  principally  as  follows  : 

In  the  first  place,  it  was  thought  adviseable  to  pubhsh  the 
work  in  two  volumes  instead  of  three  ;  the  two  volumes  being 
still  of  a  convenient  size  for  the  use  of  students. 

Secondly,  a  new  arrangement  of  various  parts  of  the  work 
has  been  adopted  Several  parts  of  the  third  volume  of  the 
English  edition  treated  of  subjects  already  discussed  in  the 
preceding  volumes  ;  in  such  cases,  when  it  was  practicable, 
the  additions  in  the  third  volume  have  been  properly  incor- 
porated with  the  corresponding  subjects  that  preceded  them  ; 

Vol.1,  }  and. 


X  PREFACE. 

and,  in  general,  such  a  disposition  of  the  various  departments 
of  the  work  has  been  made  as  seemed  best  calculated  to  pro- 
mote the  improvement  of  the  pupil,  and  exhibit  the  respec- 
tive places  of  the  various  branches  in  the  scale  of  science. 

And  thirdly,  several  notes  have  been  added  ;  and  numerous 
corrections  have  been  made  in  various  places  of  the  work  : 
it  were  tedious  and  unnecessary  to  enumerate  all  these  at  pre- 
sent ;  it  may  suffice  to  remark  the  few  following  : 

In  pages  169,  and  263,  vol.  1,  are  given  useful  notes  res- 
pecting the  degree  of  accuracy  resulting  from  the  application 
of  logarithms  ; — these  notes  will  appear  the  more  necessary  to 
beginners,  when  we  observe  such  oversights  committed  by 
authors  of  experience. 

In  page  173,  vol.  1.  a  new  definition  of  surds  is  given,  in- 
stead of  that  by  the  author  of  the  work. 

In  the  Enghsh  edition,  a  surd  is  defined  to  be  "  that  which 
has  not  an  exact  root."  In  Bonnycastle's  Algebra,  it  is  "  that 
which  has  no  exact  root."  And  in  Emerson's  Algebra,  it  is 
*■'  a  quantity  that  has  not  a  proper  root."  But  notwithstand- 
ing the  weight  of  authority  thus  evidently  against  me,  I  do  not 
hesitate  to  assert,  that  the  definition,  just  stated,  is  altogether 
erroneous.  According  to  their  definition,  the  integer  2  is  a 
surd,  for  it  "  has  not  an  exact  root." 

In  the  mensuration,  page  411,  vol.  i,  a  remark  is  added  re- 
specting the  magnitude  of  the  earth.  Dr.  Hutton  has  com- 
monly used  a  diameter  of  7957f  English  miles,  merely  be- 
cause it  gives  the  round  number  25,000  for  the  circumfe- 
rence :  in  a  few  places  he  has  used  a  diameter  of  7930.  Hav- 
ing some  years  ago  discovered  the  proper  method  of  ascer- 
taining the  most  probable  magnitude  and  figure  of  the  earth, 
from  the  admeasurement  of  several  degrees  of  the  meridian, 
I  found  the  ratio  of  the  axis  to  the  equatorial  diameter,  to  be 
as  320  to  321,  and  the  diameter,  when  the  earth  is  considered 
as  a  globe,  to  be  79187  English  miles. 

In  the  additions  immediately  preceding  the  Table  of  Loga- 
rithms in  the  second  volume,  a  new  method  is  given  for  ascer- 
taining the  vibrations  of  a  variable  pendulum.  This  problem 
was  solved  by  Dr.  Hutton,  in  his  Select  Exercises,  1787,  and 
he  has  given  the  same  solution  in  the  present  work,  see  page 
537,  vol.  2.  The  method  used  by  the  Doctor  appears  to  me 
to  be  erroneous  ;  but  in  order  that  such  as  would  judge  for 
themselves  on  this  abstruse  question,  may  have  a  fair  oppor- 
tunity of  deciding  between  us,  the  Doctor's  solution  is  given 
as  well  as  my  own. 

It  may  be  proper  to  observe,  with  respect  to  the  new  solu- 
tion, as  well  as  Dr.  Hutton's,  that  the  resulting  formula  does 

not 


PREFACE. 

not  shew  the  relation  between  the  time  and  any  number  of 
vibrations  ac<wa%  performed  ;  but  merely  gives  the  limit  to 
which  this  relation  approaches,  when  the  horizontal  velocity 
is  indefinitely  diminished.  If  therefore  we  would  use  the 
new  formula  as  slii  approximation  in  very  small  finite  vibra- 
tions, the  times  must  not  be  extended  without  limitation. 

ROBERT  ADRAIN 

JVevf-Brunraick,  JVew-Jergei/, 
July  31,  1812. 


CONTENTS 


CONTENTS 

OF  VOLUME  L 


GEJ^ERAL  Pnncipkif     .1       .       .       .       .      ^       .       .  i 

ARITHMETIC. 

JSTotation  and  Mtmeration 4, 

Jtoman  JVotation          .        .' 7 

Addition               .        .       .       .       4 .      .       .        .       .        .       .  g 

Subtraotion n 

Multiplication 13 

Division               »       •       .       ..  18 

jReduction .  gS 

Compound  Addition     .        .        .        .        .        ,        ,        .        ,       \  33 

>'     .            Subtraction .        .        .  36 

— — Multiplication /      .       ,       .        .  38 

'■■■               Division               ,       .        .      ,,       ,        ,       ;        .        ,  41 

Golden  Ruley  or  Rule  of  Three          .......  4^ 

Compound  Proportion          .        .' 49 

Vulgar  Fractions         ...........  51 

Reduction  of  Vulgar  Fractions          ,        .        ,        .        .        .        .  52 

Addition  of  Vuigar  Fractions            .......  61 

Subtraction  of  Vulgar  Fractions         ^ 62 

Multiplication  of  Vulgar  Fractions            .......  63 

Division  of  Vulgar  Fractions  ,64 

Rule  of  Three  in  Vulgar  Fractions           .       .        .        .        .       .  65 

jpecimal  Fractions               .       .        .        .       .        .        ,       .        .  66 

Ad^tion  of  Decimals 67 

l^ufttraction  Qf  Decimals 68 

Multiplication 


CONTENTS.  xiii 

Page 

Multiplication  of  Decimals      ,        .        .       .       ^        .       •       .  68 

Division  of  Decimals 70 

Reduction  of  Decimals 73 

Ilule  of  Three  in  Decimals       ........  76 

Duodecimals 77 

Involution           ...........  78 

Evolution «0 

To  extract  the  Square  Root             . 81 

To  extract  the  Cube  Root                 .        .        .         .        .        .        .  '85 

To  extract  any  Root -whatever          .......  88 

Table  of  Squares^  Cubes,  Square-Roots  and  Cube-Roots             .  90 

Ratios,  Proportions,  and  Progressions             .       .        ,        .       .  ItO 

Arithmetical  Progression          .        . Ill 

Geometrical  Progression           . 116 

Musical  Proportion            .        ,        .        .        r       ,        .        .        .  119 

Fello-wship,  ot  Partnership »1>. 

Single  Fellowship               .        ..'     .        , 120 

Double  Felloruship             .,.....-.  122 

Simple  Interest 124 

Compound  Interest             .       .        .        -        .        .      ,.        .        ,  127 

AUigation  Medial              .........  129 

AlUgation  Alternate          .        .        .        -        .        .        .        .        .  131 

Single  Position 13S 

Double  Position         ...,.,....  137 

Permutations  and  Combinations        ^               140 

Practical  Questions           .        .       .        .        .        .       ...  150 

LOGARITHMS. 

Of  Logarithms 155 

To  compute  Logarithms            . 159 

Description  and  use  of  the  Table  of  Logarithms            ...  163' 

MultipUcatim 


xiY  CONTENTS. 

Page 

Multiplication  hy  Logarithms          .......  1^7 

Division  by  Logarithms             .....««.  168 

hwolutim  by  Logarithms          . 169 

Evolution  by  Logarithms 170 

ALGEBRA. 

JDefinitions  and  Jactation          .        •        •       •       -        .       .       .  171 

.Addition              . .        .        .  17S 

Subtraction t        .        .        .        .  180 

Multiplication 181 

IHvision             ...        .        .        .       .        .       .        .        .  t^ 

Jilgebraic  Fractions 18S 

Involution 199 

Evolution           .       .               .       ,       .       .        .    '   .       .       .  202 

Surds '...,.,  205 

Infinite  Series 213 

Arithmetical  Proportion .  218 

PiUngofBaUs ,   .       .       .       .    ,     223 

Geometrical  Proportion                     .       4.1^^  '.       1      •.       ,       ,  228 

Simple  Equations 230 

Quadratic  Equations         .       .       .        *       •       •        «       .       .  24d 

JResolution  of  Cubic  and  Sigher  Equations     .       ,       .        ♦,      •  257 

Simple  Interest                                 .        .        •        •  ,«  •        •        •  266 

Compound  Interest            .       .  ^ ^    •       .  267 

Annuities           ... 270 

GEOMETRY. 

Definitions .  275 

Axioms— IVieorems            281 

Of  Ratios  and  Proportions              .%     •       •       •       •       -       *  319 

Of  Planes  and  SoHds^Dqfimtions          .       .       .       .       .       .  SS^ 

Problems  .       . .353 

Applications 


CONTENTS.  XV 

Page 

JippKcatioru  of  Mgebra  to  Geometry  369 

Plane  Trigonometry  377 

Heights  and  Diatancea  39S 

Mensuraium  of  Planes  402 

Mensuration  of  Solids  .  .      > 419 

Land  Surveying  .        •        .        -        -        -        .        .        •        .  429 

Artijiceri  Works .  487 

Timber  measuring 466 

Conic  Sections  .  .        •' 469 

Of  the  Ellipse 475 

Of  the  Hyperbola         ..........  49X 

Of  the  Parabola         ..........  514 

Of  the  Conic  Sections  as  expressed  by  Algebraic  equations,  called  the 

Equations  of  the  Curve  .        . 532 

Elements  of  Isoperimetry ,  535  . 

Problems  relative  to  the  Division  of  Surfaces  •        .       .  558 

Construction  of  Geometrical  problems 571 

Practical  Exercises  fn  M^suration  .        .        ...  575 


COURSE 


MATHEMATICS,  &c. 


GENERAL  PRINCIPLES. 

1.  Iq^UANTITY,  or  Magnitude,  is  any  thing  that  will 
admit  of  increase  or  decrease  ;  or  that  is  capable  of  any  sort 
of  calculation  or  mensuration  :  such  as  numbers,  lines,  space, 
time,  motion,  weight. 

2.  Mathematics  is  the  science  which  treats  of  all  kinds 
|of  quantity 'whatever,  that  can  be  numbered  or  meaeured. — 

That  part  which  treats  of  numt)ering  is  called  Arithmetic; 
and  that  which  concerns  measuring,  or  figured  extension,  is 
called  Geometry. — These  two,  which  are  conversant  about 
multitude  and  magnitude,  being  the  foundation  of  all  the  other 
parts,  arc  called  Pure  or  Abstract  Mathematics ;  because  they 
investigate  and  demonstrate  the  properties  of  abstract  num- 
bers and  magnitudes  of  all  sorts.  And  when  these  two  parts 
are  applied  to  particular  or  practical  subjects,  they  constitute 
the  branches  or  parts  called  Mixed  Mathematics. — Mathematics 
is  also  distinguished  into  Speculative  and  Practical :  viz.  Specu- 
lative, when  it  is  concerned  in  discovering  properties  and  re- 
lations ;  and  Practical,  when  applied  to  practice  and  real  use 
concerning  physical  objects.  ^ 

Vol.  L  2  3.  In 


2         '  GENERAL  PRINCIPLES. 

3.  In  Mathematics  are  several  general  terms  or  principles  ; 
such  as,  Definitions,  Axioms,  Propositions,  Theorems,  Prob- 
lems, Lemmas,  Corollaries,  Sclioliums,  &c. 

4.  A  Definition  is  the  explication  of  any  term  or  word  in  a 
science  ;  showing  the  sense  and  meaning  in  which  the  term 
is  employed. — Every  Definition  ought  to  be  clear,  and  ex- 
pressed in  words  that  are  common  and  perfectly  well  under- 
stood. 

6.  A  Proposition  is  something  proposed  to  be  proved,  or 
something  required  to  be  done  ;  and  is  accordingly  either  a 
Theorem  or  a  Problem. 

6  A  Theorem  is  a  demonstrative  proposition  ;  in  which 
some  property  is  asserted,  and  the  truth  of  il  required  to  be 
proved.  Thus,  when  it  is  said  that,  The  sum  of  the  three 
angles  of  any  triangle  is  equal  to  two  right  angles,  this  is  a 
Theorem,  the  truth  of  which  is  demonstrated  by  Geometry.  " 
— A  set  or  collection  of  «uch  Theorems   constitutes  a  Theory. 

7.  A  Problem  is  a  proposition  or  a  question  requiring 
something  to  be  done  ;  either  to  investigate  some  truth  or 
property,  or  to  perform  some  operation.  As,  to  find  out  the 
quantity  or  sum  of  all  the  three  anglps  of  any  triangle,  or  to 
draw  one  line  perpendicular  to  another. A  Limited  Prob- 
lem is  that  which  has  but  one  answer  or  soluiion.  An  Un-  ' 
limited  Problem  is  that  which  has  innumerable  answers. 
And  a  Determinate  Problem  is  that  which  has  a  certain  num- 
ber of  answers. 

8.  Solution  of  a  Problem,  is  the  resolution  or  answer  given 
to  it.  A  Numerical  or  Numeral  Solution,  is  the  answer©,  given 
in  numbers.  A  Geometrical  Solution^  ii  the  answer  given  by 
the  principles  of  Geometry.  And  a  Mechanical  Solution,  is 
one  which  is  gained  by  trials. 

9.  A   Lemma   is   a   preparatory  proposition,  laid   down   in 
order  to  shorten  the  demonstration  of  the  main  proposition^ 
which  follows  it.  ^P 

10.  A  Corollary,  or 'Consectary,  is  a  consequence  drawn  im- 
mediately from  some  proposition  or  other  premises.      " 

li.  A  Scholium  is  a  remark  or  observation  made  on  some 
foregoing  proposition  or  premises.  - 

12.  An  Axiom  or  Maxim^  is  a  self-evident  proposition  ; 
requiring  no  formal  demonstration  to  prove  the  truth  of  it ; 
but  is  received  and  assented  to  as  soon  as  mentioned.  Such 
as,  The  whole  of  any  thing  is  greater  than  a  part  of  it ;  or, 
The  v/hole  is  equal  to  all  its  parts  taken  together  :  or.  Two 
quantities  that  are  each  of  them  equal  to  a  third  quantity, 
are  equal  to  each  other. 

13.^ 


9  GENERAL  PRINCIPLES.  3 

13.  A  Postulate,  or  Petition,  is  something  required  to  be 
done,  which  is  so  easy  and  evident  that  no  person  will  hesi- 
tate to  allow  it. 

14.  An  Hypothesis  is  a  supposition  assumed  to  be  true,  in 
order  to  argue  from,  or  to  found  upon  it  the  reasoning^and 
demonstration  of  some  proposition. 

15.  Demonstration  is  the  collecting  the  several  arguments 
and  proofs,  and  laying  them  together  in  proper  order,  to 
show  the  truth  of  the  proposition  under  consideration. 

16.  A  Direct,  Positive,  or  Jlffirmative  Demonstration,  is 
that  which  concludes  with  the  direct  and  certain  proof  of  the 
proposit^n  in  hand, — This  kind  of  Demonstration  is  most 
satisfacWPy  to  the  mind  ;  for  which-  reason  it  is  called  some- 
times an  Ostensive  Demonstration. 

17.  Jin  Indirect,  or  Negative  Demonstration,  is  that  which 
shows  a  proposition  to  be  true,  by  proving  that  some  absur- 
dity would  necessarily  follow  if  the  proposition  advanced  were 
false.  This  is  also  sometimes  called  Reductio  ad  Ahsiirdum; 
because  it  shows  the  absurdity  and  falsehood  of  all  supposir 
tions  contrary  to  that  contained  in  the  proposition. 

18.  Method  is  the  ait  of  disposing  a  train  of  arguoients  in 
a  proper  order,  to  investigate  eithsr  the  truth  or  falsity  of  a 
proposition,  or  to  demonstrate  it  to  others  when  it  has  been 
fcund  out. — This  is  either  Analytical  or  Synthetical. 

19.  Analysis,  or  the  Analytic  Method,  is  the  art  or  mode 
of  finding  out  the  truth  of  a  'proposition,  by  first  supposing 
the  thing  to  be  done,  and  then  reasoning  back,  step  by  step, 
till  we  arrive  at  some  known  truth. — This  is  also  called  the 
Method  of  Invention,  or  Resolution. 

20.  Synthesis,  or  •  the  Synthetic  Method,  is  the  searching 
out  truth,  by  first  laying  down  some  simple  and  easy  princi- 
ples, and  pursuing  the  consequences  flowing  from  them  till 
we  arrive  at  the  conclusion. — This  is  also  called  the  Method 
of  Composition ;  and  is  the  reverse  of  the  Anal3^tic  method,  as 
this  proceeds  from  known  principles  to  an  unknown  conclu- 
sion ;  while  the  other  goes  in  a  retrograde  order,  from  the 
thing  sought,  considered  as  if  it  were  true,  to  some  known 
principle  or  fact.  And  therefore,  when  any  truth  has  been 
found  out  by  the  Analytic  method,  it  may  be  demonstrated 
by  a  process  in  the  contrary  order,  by  Synthesis. 


ARITH- 


[4] 


Arithmetic. 


jfl^RITHMETIC  is  the  art  or  science  of  numbering;  be- 
ing that  branch  of  Mathematics  which  treats  of  the  nature 
and  properties  of  numbers. — ^Wheu  it  treats  of  whole  num- 
bers, it  is  called  Vulgar,  or  Common  Arithmetic ;  but  when  of 
broken  numbers,  or  parts  of  numbers,  it  is  called  FVactions. 

Unity,  or  an  Unit^  is  that  by  which  every  thinglp  called 
one  ;  being  the  beginning  of  number  ;  as,  one  man,  one  ball, 
one  gun, 

Number  is  either  simply  one;  or  a  compound  of  several 
units  ;  as,  one  man,  three  men,  ten  men. 

An  Integer^  or  Whole  Number,  is  some  certain  precise 
quantity  of  units  ;  as,  one,  three,  ten. — These  are  so  called  as 
distinguished  from  Fractions,  which  are  broken  numbers,  or* 
parts  of  numbers  ;  as,  one-half,  two-thirds,  or  three-fourths. 


NOTATION  AND  NUMERATION. 


Notation,  or  Numeration,  teaches  to  denote  or  express 
any  proposed  number,  either  by  words  or  characters  ;  or  to 
read  and  write  down  any  sum  or  number 

The  numbers  in  Arithmetic  are  expressed  by  the  following 
ten  digits,  or  Arabic  numeral  figures,  which  were  introduced 
into  Europe  by  the  Moors,  about  eight  or  nine  hundred 
years  since  ;  viz.  1  one,  2  two,  3  three,  4  four,  5  five,  6  six, 
7  seven,  8  eight,  9  nine,  0  cipher,  or  nothing.  These  cha-^ 
racters  or  figures  were  formerly  all  called  by  the  general 
name  of  Ciphers;  whence  it  came  to  pass  that  the  art  of 
Arithmetic  was  then  often  called  Ciphering.  Also  the  first 
nine  are  called  Significant  Figures,  as  distinguished  from  the 
cipher,  which  is  of  itself  quite  insignificant. 

Besides  this  value  of  thosp  figures,  they  have  also  another 
which  depends  on  the  place  they  stand  in  when  joined  toge- 
ther ;  as  in  the  following  table  : 

Units 


NOTATION  AND  NUMERATION. 


^ 

a 
o 

H 

^ 

o 

i 

o 

03 

o 

O 
m 

1 

9 

n3 

'TO 

a 

c 

a 

C 

O 

a 

a 

u 

9 

<v 

a 

V 

ja 

s 

4) 

*2 

-a 

S 

H 

^ 

K 

H 

H 

S 

H 

t) 

&C. 

9 

8 

7 

6 

6 

4 

3 

2 

1 

9 

8 

7 

6 

6 

4 

3 

2 

9 

8 

7 

6 

5 

4 

3 

9 

8 

7 

6 

6 

4 

9 

8 

9 

7 
8 
9 

6 
7 
8 
9 

6 

6 
7 
8 
9 

Here  any  figure  in  the  fir«t  place,  reckoning  from  right  to 
left,  denotes  only  its  own  simple  value  ;  but  that  in  the 
second  place,  denotes  ten  times  its  simple  value  ;  and  that  in 
the  third  place,  a  hundred  times  its  simple  value  ;  and  so  on  : 
the  value  of  any  figure,  in  each  successive  place  being  always 
ten  times  its  former  value. 

Thus,  in  the  number  1796,  the  6  in  the  first  place  denotes 
only  six  units,  or  simply  six  ;  9  in  the  second  place  signifies 
nine  tens,  or  ninety  ;  7  in  the  third  place,  seven  hundred  ; 
and  the  1  in  the  fourth'  place,  one  thousand  :  so  that  the 
whole  number  is  read  thus,  one  thousand  seven  hundred  and 
ninety-six. 

As  to  the  cipher,  0,  though  it  signify  nothing  of  itself,  yet 
being  joined  on  the  right-hand  side  to  other  figures,  it  in- 
creases their  value  in  the  same  ten-fold  proportion  ;  thus,  5  , 
signifies  only  five  ;  but  60  denotes  5  tens,  or  fifty  ;  and  600  is 
five  hundred  ;  and  so  on. 

For  the  more  easily  reading  of  large  numbers,  they  are 
divided  into  periods  and  half-periods,  each  half-period  con- 
sisting of  three  figures ;  the  name  of  the  first  period  being 
units  ;  of  the  second,  millions  ;  of  the  third,  milUons  of 
millions,  or  bi-millions,  contracted  to  biUions  :  of  the  fourth, 
millions  of  millions  of  millions,  or  tri-roillions,  contracted  to 
trillions,  and  so  on.  Also  the  first  part  of  any  period  is  so 
many  units  of  it,  and  the  latter  part  so  many  thousands. 

The 


3  ARITHMETIC. 

The  following  Table  contains  a  summary  of  the  whole 
doctrine. 


Periods. 

Quadrill  ;  Trillions:  Billions  ;  Millions  ;  Units 

Half-per. 

th.  un.      th.  un.      th.  un.      th.  un.      th.  un. 

Figures. 

123,456  ;  789,098  ;  765,432  ;  101,234  :  567,«90. 

NuMBRATioN  is  the  reading  of  any  number  in  words 
that  is  proposed  or  set  down  in  figures  ;  which  will  be  easily 
done  by  help  of  the  following  rule,  deduced  from  the  fore- 
going tablets  and  observations — viz. 

Divide  the  figures  in  the  proposed  number,  as  in  the  sum- 
mary above,  into  periods  and  half  periods  ;  then  begin  at  the 
left-hand  side,  and  read  the  figures  with  the  names  set  to 
them  in  the  two  foregoing  tables. 

EXAMPLES. 

Express  in  words  the  following  numbers  ;  viz. 

34  15080  13405670 

96  72003  47050023 

180  109026  309025600 

304  483500  4723507689 

6134  2500639  274866390000 

9028  7523000  6578600307024 

Notation,  is  the  setting  down  in  figures  any  number  pro- 
posed in  words  ;  which  is  done  by  setting  down  the  figures 
instead  of  the  words  or  names  belonging  to  them  in  the  sum- 
mary above  ;  supplying  the  vacant  places  with  ciphers  where 
any  words  do  not  occur. 

EXAMPLES. 

Set  down  in  figures  the  following  numbers  j 
Fifty-seven. 

Two  hundred  eighty  six. 
Nine  thousand  two  hundred  and  ten. 
Twenty-seven  thousand  five  hundred  and  ninety-four. 
Six  hundred  and  forty  thousand,  four  hundred  and  eighty  one. 
Three  millions,  two  hundred  sixty  thousand,  one  hundred 

and  six. 

Four 


NOTATION  AND  NUMERATION, 


Four  hundred  and  eight  millions,  two  hundred  and  fifty-five 
thousand,  one  hundred  and  ninety-two. 

Twenty- seven  thousand  and  eight  millions,  ninety «8ix  thou- 
sand two  hundred  and  four. 

Two  hundred  thousand  and  fire  hundred  and  fifty  millions, 
one  hundred  and  ten  thousand,  and  sixteen. 

Twenty-one  billions,  eight  hundred  and  ten  millions,  sixty- 
four  thousand,  one  hundred  and  fifty. 

Of  the  Roman  Notation. 

The  Romans,  like  several  other  nations,  expressed  their 
numbers  by  certain  letters  of  the  alphabet.  The  Romans 
used  only  seven  numeral  letters,  being  the  seven  following 
capitals  :  viz.  I  for  one  ;  V  for  ^^ve  ;  X  for  ten ;  L  for  Jifty  ; 
C  for  an  hundred  ;  D  for  Jive  hundred ;  M  for  a  thousand. 
The  other  numbers  they  expressed  by  various  repetitions  and 
combinations  of  these,  after  the  following  manner  : 
1=1 

2  =  II  As  often  as  any  character  is  re- 

3  =  III  peated,  so  many  times  is  its 

value  repeated. 

4  =  nil  or  IV  A  less  character  before  a  great- 
fj  =  V  er  diminishes  its  value. 

6  =  VI  A  less  character  after  a  greater 

7  =  VII  increases  its  value. 
-      8  =  VIII 

9  =  IX 

10  =  x 

60  =  L 
100  =  C 
500  =  D  or  10 

1000  =  M  or  CID 
2000  =  MM 


5000  =  V_or  lOD 
6000  =  VI 
10000  =  X  or  CCIOD 
50000  =  Lor  1030 
60000  =  LX 

100000  =  6  or  CCCIOOO 
1000000  =  M  or  CCCCIOOOO 
2000000  =  MM 
&c.  kc. 


For  every  0  annexed,  this  be- 
comes 10  times  as  ma^y. 

For  every  C  and  0,  placed  one 
at  each  end,  it  becomes  10 
Jimes  as  much. 

A  bar  over  any  number  in- 
creases it  1000  fold. 


BtfLxr- 


ARITHMETIC. 


Explanation  op  certain  Characters. 

There  are  various  characters  or  marks  used  in  Arithmetic, 
and  Algebra,  to  denote  several  of  the  operations  and  proposi- 
tions ;  the  chief  of  which  are  as  follows  : 

-f-  signifies  plus,  or  addition. 
—     -     -     minus,  or  subtraction, 
X  or  .  -    multipUcation. 
-f-     -     -     division. 
:  :  :  :     -     proportion. 
=     -     -     equality. 
^     -     '     square  root. 
^     -     -     cube  root,  &c. 

CO     -    -     diff.  between  two  numbers  when  it  is  not  known 
which  is  the  greater. 


Thus, 

6  4-3,  denotes  that  3  is  to  be  added  to  5. 

6  —  2,  denotes  that  2  is  to  be  taken  from  6. 

7  X    3,  or  7  .  3,  denotes  that  7  is  to  be  multiplied  by  3.' 

8  -T-  4,  denotes  that  8  is  to  be  divided  by  4. 
2  :  3  :  :  4  :  6,  shows  that  2  is  to  3  as  4  is  to  6. 

6  -I-  4  =  10,  shows  that  the  sum  of  6  and  4  is  equal  to  10. 

,J  3,  or  32 ,  denotes  the  square  root  of  the  number  3. 

^  6,  or  5^,  denotes  the  cube  root  of  the  number  5. 
7s ,  denotes  that  the  number  7  is  to  be  squared. 
8 3,  denotes  that  the  number  8  is  to  be  cubed. 
&c. 


OP  ADDITION. 

Addition  is  the  collecting  or  putting  of  several  numberi 
together,  in  order  to  find  their  sum,  or  the  total  amount  of  the 
whole.    This  is  done  as  follows  : 

Set  or  place  the  numbers  under  each  other,  so  that  each 
figure  may  stand  exactly  under  the  figures  ef  the  same  value, 

that 


ADDITION.  9 

that  is,  units  under  units,  tens  under  tens,  hundreds  under 
hundreds,  &c.  and  draw  a  line  under  the  lowest  number,  to 
separate  the  given  numbers  from  their  sum,  when  it  is  found. 
— Then  add  up  the  figures  in  the  column  or  row  of  units, 
and  find  how  many  tens  are  contained  in  that  sum. — Set 
down  exactly  below  what  remains  more  than  those  tens,  or 
if  nothing  remains,  a  cipher,  and  carry  as  many  ones  to  the 
next  row  as  there  are  tens. — Next  add  up  the  second  row, 
together  with  the  number  carried,  in  the  same  manner  as  the 
first.  And  thus  proceed  till  "the  whole  is  finished,  setting 
down  the  total  amount  of  the  last  row. 


,  To  PROVE  Addition. 

First  Method. — Begin  at  the  top,  and  add  together  all  the 
rows  of  numbers  downwards  ;  in  the  same  manner  as  they 
were  before  added  upwards  ;  then  if  the  two  sums  agree,  it 
may  be  presumed  the  work  is  right. — This  method  of  proof 
is  only  doing  the  same  work  twice  over,  a  little  varied. 

Second  Method. — Draw  a  line  below  the  uppermost  number, 
and  suppose  it  cut  oflf.-^Then  add  all  the  rest  of  the  numbers 
together  in  the  usual  way,  and  set  their  sum  under  the  num- 
ber to  be  proved. — Lastly,  add  this  last  found  number 
and  the  uppermost  line  together  ,*  then  if  their  sum  be  the 
same  as  that  found  by  the  first  addition,  it  may  be  presumed 
the  work  is  right. — This  method  of  proof  is  founded  on  the 
plain  axiom,  that  *'  The  whole  is  equal  to  all  its  parts  taken 
together." 

Third  Method. — Add  the  figures  in 
the  uppermost  line  together,  and  find 
how  many  nines  are  contained  in 
their  sum. — Reject  those  nines,  and 
set  down  the  remainder  towards  the 
right-hand  directly  even  with  the 
figures  in  the  line,  as  in  the  annexed 
example. — Do  the  same  with  each 
of  the  proposedHnes  of  numbers,  set- 
ting all  these  excesses  of  nines  in  a  co-  W 
lumn  on  the  right-hand,  as  here  5,  5,  6.  Then,  if  the  excess 
of  9's  in  this  sum,  found  as  before,  be  equal  to  the  excess 
of  9's  in  the  total  sum  18304,  the  work  is  probably  right. — 
Thus,  the  sum  of  the  right-hand  column,  5,  5,  6,  is  16,  the 
excess  of  which  above  9  is  7.     Also  the  sum  of  the  figures  in 

Vol.  L  3  the 


EXAMPLE  I, 

3497 

S 

5 

6512 

a 

6 

8293 

"a 

6 

18304 

7 

O 

10 


ARITHMETIC. 


the  sum  total  18304,  is   16,  the  excess  of  which  above  9  is 
also  7,  the  same  as  the  former.* . 


OTHER  EXAMPLES. 


2. 
12345 

3. 
12345 

67890 

9876 

543 

21 

9 

4. 
12345 

67890 
98765 
43210 
12345 
67890 

876 

9087 

66 

234 
1012 

302445 

90684 

23610 

290100 

78339 

11265   " 

302445  ^ 

90684 

23610 

*  This  method  of  proof  depends  on  a  property  of  the  number  9, 
■which  except  the  number  3,  belongs  to  no  other  digit  whatever  j 
namely,  that  *'  any  number  divided  by  9,  nvUI  leave  the  same  remainder 
as  the  sum  of  its  figures  or  digits  divided  by  9  :'»  -which  may  be 
demonstrated  in  this  manner. 

'  Demonstration,  Let  there  be  any  number  proposed,  a»  4658.  This, 
separated  into  its  several  parts,  becomes  4000  •+  600  -f.50  -^  8.  But 
4000  =  4  X  1000  =-  4  X  (999  +  1)  =  4  X  999  +  4.  In  Uke  man- 
ner 600  =  6  X  99  +  6  i  and  50  ==  6  X  9  +  5.  .  herefore  the  given 
number  4658  =  4x999-|-4+6x99  +  6  +  5x9  +  5-f.  3'=4 
X  999 +  6X  99 +5x9+4  +  6+5  +  8;  and  4658  -r-  9  =  (4  x 
999  4-  6  X  99  +  5  X  9  +  4  +  6  +  5  +  8)  +•  9.  But  4  x  ^99  +  6 
X  99  +  5  X  9  is  evidently  divisible  by  9,  without  a  remainder  j  there- 
fore if  the  given  number  4658  be  divided  by  nine,  it  will  leave  the 
same  remainder  as  4  +  6+5  +  8  divided  by  9*  And  the  same,  it  is 
evident,  will  htld  for  any  other  number  whatever. 

In  like  manner,  the  san>e  property  may  be  shown  to  belong  to  the 
number  3  ;  but  the  preference  is  usually  given  lo  the  number  9,  on 
account  of  its  being  more  convenient  in  practice. 

Now,  from  the  demonstration  above  given,  the  reason  of  the  rule 
itself  is  evident :  for  the  excess  of  9*s  in  two  or  more  number^  being 
taken  separattly,  and  the  excess  of  9*s  taken  also  out  of  the  sum  of 
the  former  excesses,  it  is  plain  that  this  last  excess  must  be  equal  to 
the  excess  of  9's  contained  in  the  total  sum  of  all  these  numbers  ;  all 
the  parts  taken  together  being  equal  to  the  whole.— —This  rule 
was  first  given  by  Docter  WalUs  in  his  Arithmetici  pubUshed  in  the 
year  1657. 


Ex- 


SUBTRACTION.  11 

Ex.  5.  Add  3426 ;  9024  ;  6106  ;  8890  ;  1204,  together. 

Ans.  27650. 

6.  Add  609267  ;  236809  ;  72920  ;  83^2  ;  420  ;  21  ;  and  9, 
together.  Ans  826838. 

7.  Add  2  ;  19  ;  817  ;  4298  ;  60916  ;  730206;  91^>0634, 
together.  Ans.  9966891. 

8.  Hdw  many  days  are  in  the  twelve  calendar  months  ? 

Ans.  365. 

9.  How  many  days  are  there  from  the  1 5th  day  of  April  to 
the  24th  day  of  November,  both  days  included  ?        Ans  224. 

10.  An  army  consisting  of  62714  infantry*,  or  foot,  6110 
horse,  6250  dragoons,  3927  light-horse,  928  artillery,  or 
gunners,  1410  pioneers,  250  sappers,  and  406  miners  :  what 
is  the  whole  number  of  men  ?  Ans.  70995. 


OF  SUBTRACTION. 


Subtraction  teaches  to  find  how  much  one  number 
exceeds  another,  called  their  difference,  or  the  remainder^  by 
taking  the  less  from  the  greater.  The  method  of  doing  which 
is  as  follows  : 

Place  the  less  number  under  the  greater,  in  the  same  man- 
ner as  in  Addition,  that  is,  units  under  units,  tens  under  tens, 
and  so  on  ;  and" draw  a  line  below  them. — Begin  at  the  right- 
hand  and  take  each  figure  in  the  lower  line,  or  number,  from 
the  figure  above  it,  setting  down  the  remainder  below  it. — 
But  if  the  figure  in  the  lower  line  be  greater  than  that  above 
it,  first  borrow,  or  add,  10  to  the  upper  one,  and  then  take 
the  lower  figure  from  that  sum,  setting  down  the  remainder^ 
and  carrying  1 ,  for  what  was  borrowed,  to  the  next  lower 
figure,  with  which  proceed  as  before  ;  and  so  on  till  the 
whole  is  finished. 


*  The  whole  body  of  foot  soldiers  is  denoted  by  the  word  Infantry  ; 
and  all  those  that  charge  on  horseback  by  the  word  Cavalry.  —  Some 
authors  conjecture  tHat  the  term  infantry  is  derived  from  a  certain  In- 
fanta of  Spain,  who  finding  that  the  ariry  commanded  by  the  k'mg  her 
father  had  been  defeated  by  the  Moors,  assembled  a  body  of  the  peo- 
ple together  on  foot,  with  which  she  engaged  and  totally  routed  the 
enemy.  In  honour  of  this  event,  and  to  distinguish  the  foot  soldiers, 
who  were  not  before  held  in  much  estimation,  they  received  the  name 
of  Infantrv,  from  her  own  title  of  Infanta. 

To 


12  ARITHMETIC. 

i 

To  FROVE  Subtraction. 

Add  the  remainder  to  the  less  number,  or  that  which  is 
just  above  it ;  and  if  the  sum  be  equal  to  the  greater  or  upper- 
most number,  the  work  is  right*. 


EXAMPLES. 


From  5386427 
Take  2164315 


From  5386427 
Take  4258792 


Prom  1234567 
Take  702973 


Rem.  3222112 


Rem.  1127635 


Rem.  531594 


Proof.  5386427 


Proof.  5386427 


Proof.  1234567 


4.  From  6331896  take  5073918. 
6.  From  7020974  take  2766809. 
6.  From  8503602  take  574271. 


Ads.  25 
Ans.  4254165. 
Ans.  7929131. 


7.  Sir  Isaac  Newton  was  born  in  the  year  1642,  and  he 
clied  in  1727  :  how  old  was  he  at  the  time  of  his  decease  ? 

Ans.  85  years. 

8.  Homer  was  born  2543  years  ago,  and  Christ  1810  years 
ago  :  then  how  long  before  Christ  was  the  birth  of  Homer  ? 

Ans.  733  years. 

9.  , Noah's  flood  happened  about  the  year  of  the  world  1656, 
and  the  birth  of  Christ  about  the  year  4000  :  then  how  long 
was  the  flood  before  Christ  ?  Ans.  2344  years. 

10.  The  Arabian  or  Indian  method  of  notation  was  first 
known  in  England  about  the  year  1150:  then  how  long  is 
it  since  to  this  present  year  1810  ?  Ans.  660  years. 

11.  Gunpowder  was  invented  in  the  year  1330  :  then  how 
long  was  this  before  the  invention  of  printing,  which  was 
in  1441  ?  Ans.  Ill  years. 

12.  The  mariner's  compass  was  invented  in  Europe  in  the 
year  1302  :  then  how  long  was  that  before  the  discovery  of 
America  by  Columbus,  which  happened  in  1492  ? 

Ans.  190  years. 


*  The  reason  of  tlii?  method  of  proof  is  evident ;  for  if  the  differ- 
ence of  two  nunrjbers  be  added  to  the  less,  it  must  manifestly  make 
up  a  sum  equal  to  the  greater. 

OF 


MULTIPLICATION. 


l.*^ 


OF  MULTIPLICATION. 

Multiplication  is  a  compendious  method  of  Addition, 
teaching  how  to  find  the  amount  of  any  given  number  when 
repeated  a  certain  number  of  times;  as,  4  times  6,  which 
is  24. 

The  number  to  be  multiplied,  or  repeated,  is  called  the 

Multiplicand. — The  number  you  multiply  by,  or  the  number  of 

^^repetitions,  is  the   Multiplier. — -And  the  number  found,  being 

^     the    total  amount,    is    called   the    Product. — Also,    both  the 

multiplier  and  multiplicand  are,  in  general,  name  the  Terms 

or  Factors. 

Before  proceeding  to  any  operations  in  this  rule,  it  is  ne- 
cessary to  learn  off  very  perfectly  the  following  Table,  of  all 
the  products  of  the  first  12  numbers,  commonly  called  the 
Multiplication  Table,,  or  sometimes  Pythagoras's  Table,  from 
its  inventor. 


Multiplication  Table. 


1 

2 
3 
4 
5 
6 
~ 
8 
9 

10 

11 

12 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

3 
6 
9 
12 
15 
18 
21 
24 

4 
8 
12 
16. 
20 
24 
28 
32 
36 
40 
44 
48 

10 
15 
20 
25 
30 
35 
40 
45 
50 
55 
60 

6 
12 
18 
24 
30 
36 
42 
48 
54 
60 
66 
72 

7 
14 
21 
28 
35 
42 
49 
56 
63 
70 
77 
84 

8 
16 
24 
32 
40 
48 
56 
64 
72 
80 
S8 
96 

9 
18 
27 
36 
45 
54 
63 
72 
81 
90 
99 
108 

10 
20 
30 
40 
50 
60 
70 
80 
90 
100 
110 
120 

11 

22 
33 
44 
5.5 
66 
77 
88 
99 
110 
121 
132 

12 
24 
^6 
48 
60 
72 
84 
96 
108 
120 

27 
30 
33 
36 

132 

144 

To 


14  ARITHMETIC. 


To  multiply  any  Given  Number  by  a  Single  Figure^  or  by  any 
Number  not  more  than  12. 

*  Set  the  multiplier  under  the  units  figure,  or  right-hand 
place,  of  the  multiplicand,  and  draw  a  line  below  it. — T  hen 
beginning  at. the  right  hand,  multiply  every  figure  in  this  by 
the  multiplier. — Count  how  many  tens  there  are  in  the  pro- 
duct of  every  single  figure,  and  set  ilown  the  remainder  di- 
rectly under  the  figure  that  is  multiplied  ;  and  if  nothing 
remains,  set  down  a  cipher. — Carry  as  many  units  or  ones  as 
there  are  tens  couuted,  to  the  product  of  the  next  figures  ; 
and  proceed  in  the  same  manner  till  the  whole  is  finished. 

EXAMPLE. 

Multiply  9876543210  the  Multiplicand. 
By      -     -      -      -     2  the  Multiplier. 


19753086420  the  Product. 


To  multiply  by  a  Number  consisting  of  Several  Figures. 

t  Set  the  multiplier  below  the  multiplicand,  placing  them 
as  in  Addition,  namely,  units  under  units,  tens  under  tens,  &c. 
drawing  a  line  below  it. — Multiply  the  whole  of  the  multi- 
plicand by  each  figure  of  the  multiplier,  as  in  the  last  article; 

setting 


♦  The  reason  of  this  rule  is  the  same  as  for 
the  process  in  Addition,  in  which  1  is  car- 
ried for  every  10,  to  the  next  place,  gra- 
dually as  the  several  products  are  produced 
one  after  another,  instead  of  setting  them 
all  down  one  below  each  other,  as  in  the  an- 
nexed example. 


f  After  having  found  the  produce  of  the  multiplicand  by  the  first 
figure  of  the  multiplier,  as  in  the  former  case,  the  multiplier  is  sup- 
posed to  be  divided  into  parts,  and  the  product  is  found  for  the  second 
figure  in  the  same  manner :  but  as  this  figure  stands  in  the  place  of 
tens,  the  product  must  be  ten  times  its  simple  v.due  ;  and  therefore  the 
first  figure  of  this  product  roust  be  set  in  the  place  of  tens  j  or,  which 


5678 

4 

32  = 

8 

X 

4 

280  = 

70 

X 

4 

2400  = 

600 

X 

4 

20000  = 

5000 

X 

4 

22712  = 

5678 

X 

4 

MULTIPLICATION.  16 

setting  down  aline  of  products  for  each  figure  in  the  multi- 
plier, so  as  that  the  first  figure  of  each  line  may  stand  straight 
under  the  figure  multiplying  by. — Add  all  the  lines  of  pro- 
ducts together,  in  the  order  as  they  stand,  and  their  sum  will 
be  the  answer  or  whole  product  required. 

To  PROVE  Multiplication. 

There  are  three  different  ways  of  proving  Multiplication, 
which  are  as  below  : 

First  Method. — Make  the  multiplicand  and  multiplier 
change  places,  and  multiply  the  latter  by  the  former  in  the 
same  manner  as  before.  Then  if  the  product  found  in  this 
way  be  the  same  as  the  former,  the  number  is  right. 

Second  Method. ■^■*CdiSt  all  the  9's  out  of  the  sum  of  the 
figures  in  each  of  the  two  factors,  as  in  Addition,  and  set 
down  the  remainders.  Multiply  these  two  remainders 
together,  and  cast  the  9's  out  of  the  product,  as  also   out  of 


is  the  same  thing,  directly  under  the  figure  multiplied  by.     And  pro- 
ceeding in  this  manner  sepa 
rately  with  all  the  figures  of  ' 

the  multiplier,   it  is  evident        123456r    the  multiplicand, 
that  we  shall  multiply  all  the  4567 

parts  of  the  multiplicand  by  — — 

all   the   parts  of    the   multi-         8641969=       7  times  the  mult, 
plier,     or   the   whole  of  the  '    7407402    =     60  times  ditto, 
multiplicand    by    the    whole     6172835      ==  500  times  ditto- 
of  the  multiplier  ;    therefore  493B268        =4000  times  ditto. 

these  several  products  bemg .  ^  i. 

added  together,  will  be  equal  563826:489  =4567  times  ditto, 
to  the    whole   required  pro-         i. 
duct  ;  as  in  the  example  an- 
nexed. 

*  This  method  of  proof  is  derived  fi-om  the  peculiar  property  of  the 
number  9,  mentioned  in  the  proof  of  Addition,  and  the  reason  for  the 
one  may  serve  for  that  of  the  other.  Another  more  ample  demonstra- 
tion of  thts  rule  may  be  as  follows  :— Let  P  and  (^denote  the  number 
of  9's  in  the  factors  to  be  multiplied,  and  a  and  b  what  remain  ;  then  9 
P  4-  a  and  9  Q_"f  6  will  be  the  numbers  themselves,  and  their  product  is 
(9  P  X  9  Q)  +  (9  i"  X  6)  +  (9  ax  a)  +  (a  X  A),  but  the  first  three 
of  these  products  are  each  a  precise  number  of  9's,  because  their  fac- 
tors are  so,  either  one  or  both  :  these  therefore  being  cast  away,  there 
remains  only  aX  ^  ;  and  if  the  9's  also  be  cast  out  of  this,  the  excess 
is  the  excess  of  9's  in  the  total  product  :  but  a  and  b  are  the  ex- 
cesses in  tlie  factors  themselves,  and  a  X  b  is  their  product  ;  therefore 
the  rule  is  true. 

the 


16 


ARITHMETIC. 


the  whole  product  or  answer  of  the  question,  reserving  the 
remainders  of  these  last  two,  which  remainders  must  be  equal 
when  the  work  is  right. — Note^  It  is  common  to  set  the  foui-^ 
remainders  within  the  four  angular  spaces  of  a  cross,  as  in  the 
example  below. 

Third  Method. — Multiplication  is  also  very  naturally 
proved  by  Division  ;  for  the  product  divided  by  either  of  the 
factors,  will  evidently  give  the  other.  But  this  cannot  be 
practised  till  the  rule  of  Division  is  learned. 


Mult.  3542 
by       6196 

EXAMPLES. 

Proof. 

X 

or  Mult.  6196 
by       3542 

21252 
31878 
3542 
2125^ 

12392 
24784 
30980 
18588 

21946232  Product. 

21946232  Proof. 

OTHER  EXAMPLES. 


Multiply  123456789  by  3. 

Ans. 

370370367. 

Multiply   123456789  by  4. 

Ans. 

493827156. 

Multiply   123456789  by  5. 

Ans. 

617283945. 

Multiply   123456789  by  6. 

Ans. 

740740734. 

Multiply  123456789  by  7. 

Ans. 

864197523. 

Multiply  123456789  by  8. 

Ans. 

987654312. 

Multiply  123456789  by  9. 

Ans. 

llllIlllOl. 

Multiply  123456789  by   11. 

Ans. 

1358024679. 

Multiply  123456789  by   12. 

Ans. 

1481481468. 

Multiply  302914603  by  16. 

Ans. 

4846633648. 

Multiply  273580961   by  23. 

Ans. 

6292362103. 

Multiply  402097316  by  195. 

Ans. 

78408976620. 

Multiply  82164973     by  3027. 

Ans. 

248713373271. 

Multiply  7564900       by  579. 

Ans. 

4380077100. 

Multiply  8496427       by  874359. 

Ans. 

7428927415293. 

Multiply  2760325       by  37072. 

Ans. 

102330768400. 

CONTRAC- 


MULTIPLICATION.  17 

GONTBACTIONS  IN  MULTIPUCATION. 

L  When  there  are  Ciphers  in  the  Factors, 

If  the  ciphers  be  at  the  right-hand  of  the  numbers  ;  mul- 
tiply the  other  figures  only,  and  annex  as  many  ciphers  to 
the  right' hand  of  the  whole  product,  as  are  in  both  the  fac- 
tors.— When  the  ciphers  are  in  the  middle  parts  of  the  mul- 
tipher  ;  neglect  them  as  before,  only  taking  care  to  place 
the  first  figure  of  every  line  of  products  exactly  under  the 
figure  multiplying  with. 

EXAMPLES. 

1.  2. 

Mult.  9001635  Mult.  390720400 

by      -     gOlOO  by       -      406000 


9001635  23443224 

63011445  166^:8816 


631014613500  Products  158632482400000 


3.  Multiply  81503600  by  7030.  Ans.  572970308000. 

4.  Multiply  9030100    by  2100.  Ans.  189632»0000 
6.  Multiply  a057069    by  70050.           Ans.  564397683450. 

II.   When  the  multiplier  is  the  Product  of  two  or  more  Numbers 
in  the  Table  :  then 

*  Multiply  by  each  of  those  parts  separately,  instead  of  the 
whole  number  at  once. 

EXAMPLES, 

1.  Multiply  51307298  by  56,  or  7  times  8. 
51307298 

7 


359151086 
8 

2873208688 


♦  Tlie  reason  of  this  rule  is  obvious  enough  ;  for  any  number  mul- 
tiplied by  the  component  parts  of  another,  must  give  the  same  pro- 
duct as  if  it  were  multipUed  by  that  number  at  once.  Thus,  in  the 
1st  example,  7  times  the  product  of  8  by  the  given  number,  makes  56 
times  the  same  number,  as  plainly  as  7  times  8  makes  56. 

Vol.  I.  ,  4  2.  Mul- 


IB  ARITHMETIC* 

2.  Multiply  31704592     by  36.  Ana.  1141365312. 

3.  Multiply  29753804     by  72.  Ans.  2142273888. 

4.  Multiply  7128368       by  96.  Ans.  684323328. 
6.  Multiply  160430800  by  108.  Ans.  17326526400. 

6.  Multiply  61835720     by  1320.         Ans.  81623150400. 

7.  There  was  an  army  composed  of  104  *  battalions,  eacb 
consisting  of  500  men  ;  what  was  the  number  of  men  con- 
tained iR  the  whole  ?  Ans.  52000. 

8.  A  convoy  of  ammunition  f  bread,  consisting  of  250 
waggons,  and  each  waggon  containing  320  loaves,  having 
been  intercepted  and  taken  by  the  enemy  ;  what  is  the  num- 
ber of  loaves  lost  ?  Ans.  80000. 


OF  DIVISION. 

.• 

Division  is  a  kind  of  compendious  method  of  Subtrac- 
tion, teaching  to  find  ho\V  often  one  number  is  contained  ia 
another,  or  may  be  taken  out  of  it :  which  is  the  same  thing. 

The  number  to  be  divided  is  called  the  Dividend, — 
The  number  to  divide  by,  is  the  Divisor. — And  the  number 
of  times  the  dividend  contains  the  divisor,  is  called  the  Quo^ 
tient. — Sometimes  there  is  a  Remainder  left,  after  the  divi- 
sion is  finished. 

The  usual  manner  of  placing  the  terms,  is  the  dividend  in 
the  middle,  having  the  divisor  on  the  left  baud,  and  the  quo- 
tient on  the  right,  each  separated  by  a  curve  line ;  as,  td 
divide  12  by  4,  the  quotient  is  3, 

Dividend  12 

Divisor  4)  12  (3  Quotient ;  4  subtr. 

showing  that  the  number  4  is  3  times         — 
contained    in    12,   or   may  be   3  times  8 

subtracted  out  of  it,  as  in  the  margin.  4  subtr. 

J  Rule — Having    placed   the    divisor         — 
before  the    dividend,  as    above  direct-  4 

ed,  find  how  often  the  divisor  is  con-  4  subtr. 

tained  in  as  many  figures  of  the   divi-         — 
dend  as  are  just  necessary,  and  place  the  0  . 

number  on  the  right  in  the  quotient.  — 

Mul- 


*  A  battalion  is  a  body  of  foot,  consisting  of  500,  or  600,  or  700  men, 
more  or  less. 

\  The  ammunition  bread,  is  that  which  is  provided  for,  and  distri- 
buted to,  the  soldiers  j  the  usual  allowance  being  a  loaf  of  6  pounds  to 
every  soldier,  Once  in  4  da)'s. 

*  In  this  way  the   dividend  is  resolved  into  parts,  and  by  trial  is 

found 


DIVISION.  1» 

Multiply  the  divisor  by  this  number,  and  set  the  product 
under  the  figures  of  the  dividend  before-mentioned. — Sub- 
tract this  product  from  that  part  of  the  dividend  under  which 
it  stands,  and  bring  down  the  next  figure  of  the  dividend,  or 
more  if  necessary,  to  join  on  the  right  of  the  remainder — Di- 
vide this  number,  so  increased,  in  the  same  manner  as  before  ; 
and  so  on  till  all  the  figures  are  brought  down  and  used. 

JV.  B.  If  it  be  necessary  to  bring  down  more  figures  than 
one  to  any  remainder,  in  order  to  make  it  as  large  as  the 
divisor,  or  larger,  a  cipher  must  be  set  in  the  quotient  for 
every  figure  so  brought  down  more  than  one. 


TO  PROVE  DIVISION. 

4 

*  Multiply  the  quotient  by  the  divisor  ;  to  this  product 
add  the  remainder,  if  there  be  any  ;  then  the  sum  will  be 
equal  to  the  dividend  when  the  work  is  right. 


found  how  often  the  divisor  is  contained  in  each  of  those  parts,  one 
after  another,  arranging  the  several  figures  of  the  quotient  one  after 
another,  into  one  number. 

When  there  is  no  remainder  to  a  division,  the  quotient  is  the  whole 
and  perfect  answer  to  the  question.  But  when  there  is  a  remainder, 
it  goes  so  much  towards  another  time,  as  it  approaches  to  the  divisor  ; 
so,  if  the  remainder  be  half  thfe  divisor,  it  will  go  the  half  of  a  time 
more  ;  if  the  4th  part  of  the  divisor,  it  will  go  one  fourth  of  a  time 
more  ;  and  so  on.  Therefore,  to  complete  the  quotient,  set  the  re- 
mainder at  the  end  of  it,  above  a  small  line,  and  the  divisor  below  it 
thus  forming  a  fractional  part  of  the  whole  quotient. 

♦This  method  of  proof  is  plain  enough:  for  since  the  quotient 
is  the  number  of  times  the  dividend  contains  the  divisor,  the  quo- 
tient multiplied  by  the  divisor  must  evidently  be  equal  to  the  divi- 
dend. 

There  are  also  several  other  methods  sometimes  used  for  proving 
Division,  some  of  the  most  useful  of  which  are  as  follow  : 

Second  ikferAorf— Subtract  the  remainder  from  the  dividend  ;  and 
divide  what  is  left  by  the  quotient  ;  so  shall  the  new  quotient  from 
this  last  division  be  equal  to  the  former  divisor,  when  the  work  is 
right. 

Third  Method— '\ddtogeiher  the  remainder  and,  all  the  products  of 
the  several  quotient  figures  by  the  divisor,  according  to  the  order  in 
which  they  stand  in  the  work  ;  and  the  sum  Avill  be  equal  to  the  divi- 
dend when  the  work  is  right.  .  , 

EXAM 


2(^ 


ARITHMETIC. 

EXAMPLES. 

1. 

1234567 
12 

Quot. 
(411522 
mult.  3 

37) 

2. 
12345678 
111 

124 
111 

135 
111 

246 

222 

Q,aot. 

(333666. 

37 

3 
3 

1234566 
add  1 

2335662 
1000998 
rem.  36 

4 

1234567 

3   - 

12345678 

'  ■  ■    F 

•roof. 

15 
35 

Proof. 

6 
6 

%41 

222 

7 
6 

268 

222 

Rem.  1 


Rem.  36 


3.  Divide  73146085   by  4. 

4.  Divide  5317986027  by  7. 
6.  Divide  570196382  by  12. 

6.  Divide  7463810,6   "" 

7.  Divide  137896264 

8.  Divide  35821649 

9.  Divide  72091365 
10.  Divide  4637064283 


Ans.  182865211. 

Ans.  759712289f 

Ans  47616366f^. 

Ans.  20 1 7246 3^. 

Ans. 


Ans.  46886lff. 
Ans.   13861-i-OoV. 
Ans.  80496^VHV. 


by  37. 
by  97. 
by  764. 
by  5201. 
by  67606. 

11.  Suppose  471  men  are  formed  into  ranks  of  three  deep, 
what  is  the  number  in  each  rank  ?  Ans.  167. 

12.  A  party  at  the  distance  of  378  miles  from  the  head 
quarters,  receive  orders  to  join  the  corps  in  18  days  :  what 
number  of  miles  must  they  march  each  day  to  obey  their 
orders  ?  Ans.  21. 

13.  The  annual  revenue  of  a  gentleman  being  383.^0/; 
how  much  per  day  is  that  equivalent  to,  there  being  366  d^ys 
in  the  year  ?  Ans.    104/. 


CONTRACTIONS  IN  DIVISION. 


There  are  certain  contractions  in   Division,  by  which  the 

operation  in  particular  cases  may  be  performed  in  a  shorter 
manner  as  follows  : 

*  I.  Divi- 


DIVISION. 


21 


I.  Division  by  any  Small  Number,  not  greater  than  12,  may 
be  expeditiously  performed,  by  multiplying  and  subtracting 
mentally,  onutting  to  set  down  the  work,  except  only  the 
quotient  immediately  below  the  dividend. 


3)  66103961 
Quot.   18701320i 


EXAMPLES. 

4)  52619676         6)     1379192 


6)  38672940         7)  8 1 396627         8)  237 1 8920 


9)  43981962       11)  67614230       12)  27980373 


II.  ^When  Ciphers  are  annexed  to  the  Divisor;  cut  off  those 
ciphers  from  it,  and  cut  off  the  same  number  of  figures  from 
the  ri<.l<t-hand  of  the  dividend  :  th^en  divide  with  the  remain- 
ing tigures,  as  usual.  And  if  there  be  any  thing  remaining 
after  this  division,  place  the  figures  cut  off  from  the  dividend 
to  the  right  of  it,  and  the  whole  will  be  the  true  remainder  ; 
otherwise,  the  figures  cut  off  only  will  be  the  remainder. 


EXAMPLES. 


1.  Divide  3704196  by  20. 
.    2,0)  370419,6 


Quot.  186209^1 


2.  Divide  31086901  by  7100. 
71,00)  310869,01    (4378|i^i.. 
284 

268 
213 


656 
497 


699 
568 


31 


3.  Divide 


*  This    method  is  only  to    avoid  a  needless  repetition  of  ciphers 
which  would  happen  in  the  common  way.    And  the  truth  of  the 

principle 


22  ARITHMETIC. 

3.  Divide  7380964  by  23000.  Ans.  320f||f| 

4.  Divide  2304109  by  5800.  Ans.    397if^f . 
III.  When  the  Divisor  is  the  exact  Product  of  two  or  more 

of  the  small  Numbers  not  greater  than  12  :  *  Divide  by  each 
©f  those  numbers  separately,  instead  of  the  whole  divisor  at 
once. 

N.  B.  There  are  commonly  several  remainders  in  virork- 
ing  by  this  rule,  one  to  each  division  ;  and  to  find  the  true 
or  whole  remainder,  the  same  as  if  the  division  had  been  per- 
formed all  at  once,  proceed  as  follows  :  Multiply  the  last 
remainder  by  the  preceding  divisor,  or  last  but  one,  and  to 
the  product  add  the  preceding  remainder  ;  multiply  this  sum 
by  the  nest  preceding  divisor,  and  to  the  product  add  the 
next  preceding  remainder  ;  and  so  on^  till  you  have  gone 
backward  through  all  the  divisors  and  remainders  to  the  first. 
As  in  the  example  following  : 

EXAMPLES. 

1.  Divide  31046835  by  56  or  7  times  8. 
7)  31046835  6  the  last  rem. 

mult.  7  preced.  divisor.^ 


S)     4435262—1  first  rem. 


42 


554407 — 6  second  rem.  add        1  the  1st  rem. 


Ans.  554407f|  43  whole  rem. 


2.  Divide  7014596  by  72. 

3.  Divide  5130652  by  132.  Ans.  38868/3-. 

4.  Divide  83016572  by  240.  Ans.  345902^?^-. 

principle  on  which  it  is  founded  is  evident:  for  cutting  off  the  same 
number  of  ciphers,  or  figures,  from  each,  is  the  same  as  dividing  each 
of  them  by  10,  or  100.  or  1000,  &c.  according  to  the  number  of  ciphers 
cut  off;  and  it  is  evident,  that  as  often  as  the  whole  divisor  is  contain- 
ed in  the  whole  dividend,  so  often  must  any  part  of  the  former  be  con- 
taine(J  in  a  like  part  of  the  latter. 

*  This  follows  from  the  second  contraction  in  Multiplication,  being 
only  the  converse  of  it ;  for  the  half  of  the  third  part  of  any  thing, 
is  evidently  the  same  as  the  sixth  part  of  the  whole  ;  and  so  of  any 
other  numbers.— The  reason  of  the  method  of  finding  the  whole  re- 
mainder from  the  several  particular  pnes,  will  best  appear  from  the 
nature  of  Vuigiar  fractions.  Thus  in  the  first  example  above,  the 
first  remainder  being  1,  when  the  divisor  is  7>  makes  \  this  must  be 
added  to  the  second  remainder,  6,  making  6\  to  the  divisor  8,  or  to  be 
divided  by  8.    B»\t  6    »6x7  +  l       43 

43  43       * _=-;   and  this   divided  by  a. 

gives— =  — -  7  7 

7X8       56 

IV.  Common. 


REDUCTION.  25 

IV.  Common  Division  may  be  performed  more  concisely ^ 
by  omitting  the  several  products,  and  setting  down  only  the 
remainders  ;  namely,  multiply  the  divisor  by  the  quotient 
figures  as  before,  and  without  setting  down  the  product,  sub- 
tract each  figure  of  it  from  the  dividend,  as  it  is  produced  ; 
always  remembering  to  carry  as  many  to  the  next  figure  as 
were  borrowed  before. 

^  EXAMPLES. 

1.  Divide  3104G79  by  833. 
833)  3104679  (^12'7j%^:,. 
6056 
2257 
5919 
88 
2.  Divide  79165238  by  238,      Ans.  332627^^* 
3  Divide  29137062  by  5317.     Ans.  54791114. 
4.  Divide  62016735  by  7803.      Ans.  79474ff|. 


OF  REDUCTION. 

Redvction  is  the  changing  of  numbers  from  one  naaoe 
or  denomination  to  another  without  altering  their  -value. — 
This  is  chiefly  concerned  in  reducing  money,  weights,  and 
measures. 

When  the  numbers  are  to  be  reduced  from  a  higher  name 
to  a  lower,  it  is  called  Reduction  Descending ;  but  when, 
contrarywise,  from  a  lower  name  to  a  higher,  it  is  Reduction 
Ascending. 

Before  proceeding  to  the  rules  and  questions  of  Reduction, 
it  will  be  proper  to  set  down  the  usual  Tables  of  Moneys 
weights,  and  measures,  which  are  as  follow  : 

Of  MONEY,  WEIGHTS,  AND  MEASURES. 

TABLES  OF  MONEY.» 

2  Farthings  =  1  Halfpenny  \  \       qrs  d 

4  Farthings  =  1  Penny         d\        4=1  « 

12  Fence         =   1  Shilling       si      48  =     12  =  1  £ 

20  Shillings     =   1  Pound        ^  |     960  =  240  =s  20  =^  1 

PENCE 

♦^denotes  pounds,  s  shillings,  and  d  denotes  pence, 
i      denotes  1  farthing,  or  oiie  quarter  of  any  thing- 
■J      denotes  a  halfpenny  or  the  half  of  any  thing. 
^      denotes  3  farthings  or  three  quarters  of  any  thing. 

The 


24 


ARITHMETIC. 


PENCE  TABLE. 

d 

s     d 

20     is 

1      8 

30    ~ 

2     6 

40     — 

3     4 

60     — 

4     2 

60     — 

6     0 

70     ~ 

5  10 

80     — 

6     8 

90     — 

7     6 

100     — 

8     4 

110     — 

9     2 

120     — 

10     0 

SfflLLINGS  TABLE; 


s 

d 

1 

is 

12 

^ 

— 

24 

3 

— 

36 

4 

— 

48 

5 

— 

60 

6 

— 

72 

7 

— 

84 

8 

— 

96 

9 



108 

10 

— 

120 

11 



132 

FEDERAL  MONEY. 


10  Mills  (m)=  1  Cent  c 
10  Cents  =  1  Dime  d 
10  Dimes  =  1  Dollar  D 
10  Dollars     =  1  Eagle  E 


Standard  Weight,  dwt  gr^ 
The  Cent  weighs  6    23  Copper 
Dollar  17      ]|  Silver 

Eagle  11      4|  Gold 


The  standard  for  Federal  Money  of  Gold  and  Silver  is  1 1 
parts  fine,  and  1  part  alloy. 

A  Dollar  is  equal  to  4s  and  8d  in  South  Carolina,  to  6s  in  the 
New-England  Spates  and  Virginia,  to  7s  and  6d  in  New- 
Jersey,  Pennsylvania,  Delaware,  and  Maryland,  and  to  8s  in 
New-York,  and  North-Carohna. 

TROY 


The  full  weight  and  value  of  the  English  gold  and  silver  coin,  is  as 


here  below  : 
Gold. 


Vaiue» 
£,  s  a 
110 
0  10     6 


A  Guinea 
Half  guinea 
Seven  Shillings  0     7 
Qiiarter-guinea  0     5 


W tight. 

diut  gr 
5  9i 
2  164 
1  19| 
1     8^ 


Silver. 

Value. 

Weight. 

s     d 

dint  gr 

A  Crown 

5    0 

19     8§ 

Half-crown 

2     6 

9  16i 

Shilhng 

1    0 

3  21 

Sixpence 

0     6 

1  22i 

The  usual  value  of  gold  is  nearly  4?  an  ounce,  or  2d  a  grain  ;  and 
that  of  Silver  is  nearly  Ss  an  ounce.  Also  the  value  of  any  quantity 
of  gold,  is  lo  the  value  of  the  same  weight  of  standard  silver,  nearly 
as  15  to  1,  or  more  nearly  as  15  and  l-14th  to  1. 

Pure  gold,  free  from  mixture  with  other  metals,  usually  called 
fine   gold   is    of  so  pure   a   nature,  tliat  it    will    endure  the    fire 

without 


TABLES  OP  WEIGHTS.  26 

TROYtVElGriT> 


Grains  -  -  marked  ^r 
24  Grains  make  1  Penny  weight  rfa>< 
20  Pennyweights  1  Ounce  oz 

12  Ounces  1  Pound  lb 


gr         dwt 
24  =       I      oz 
480=     20=   1      lb 
5760=  24^=12  =  1 


By  this  weight  are  Weighed  Gold,  Silver,  and  Jewels. 
APOTHECARIES'  WEIGHT. 


Grains      - 
20  Grains  make 

3  Scruples  - 

8  Drams 
12  Ounces    - 

1 

-  1 

-  1 

-  1 

marked  gr 
Scruple  sc  or 
Dram     dr  or 
Ounce   oz  or 
Pound    lb  or 

3 

fb 

gr             sc 
20  =        1 
60  =       3  = 
480  =     24  = 
6760  =  988  = 

dr 

1 

8 

96 

oz 
=       1 
=     12 

Ih 
=     1 

This  is  the  same  as  Troy  weight,  only  having  some  differ- 
ent divisions.  Apothecaries  niake  use  of  this  weight  in  com- 
pounding their  Medicines  ;  but  tiiey  buy  and  seH  their  Drogi 
by  Avoirdupois  weight. '  Ak  ^ 


-without  wasting,  though  it  be  kept  continually  melted.  But  silver,  not 
having  the  purity  of  gold,  will  not  endure  the  fire  like  it  ;  yet  fine 
silver  will  waste  but  a  very  little  by  being  in  the  fire  any  moderate 
time  ;  whereas  copper,  tin,  lead,  &c.  will  not  only  waste,  but  may  be 
calcined,  or  burnt  to  a  powder. 

Both  gold  and  silver,  in  their  purity,  are  so  very  soft  and  flexible 
(like  new  lead,  &c),  that  they  are  not  so  useful,  either  in  coin  op 
otherwise  (except  to  beat  into  leaf  gold  or  silver),  as  when  they  are 
allayed,  or  mixed  and  hardened  with  copper  or  brass.  And  though 
most  nations  differ,  more  or  less,  in  the  quantity  of  such  allay,  as  well 
as  in  the  same  place  at  different  times,  yet  in  England  the  standard  for 
gold  and  silver  coin  has  been  for  a  long  time  as  follows — viz.  That  22 
parts  of  fine  gold,  and  2  parts  of  copper,  being  melted  together,  shall 
be  esteemed  the  true  standard  for  gold  coin  :  And  tliat  11  ounces  and 
2  pennyweights  of  fine  silver,  and  18  pennyweights  of  copper,  being 
melted  together,  is  esteemed  the  true  standard  for  silver  coin,  called 
Sterling  silver. 

*  The  original  of  all  weights  used  in  England,  was  a  grain  op 
com  of  wheat,  gathered  out  of  the  middle  of  the  ear,  and,  being 
well  dried,  32  of  them  were  to  make  one  pennyweight,  20  penny- 
weights 

Vol.  I.  5 


26 


ARITHMETIC, 


AVOIRDUPOIS  WEIGHI*. 


Drams 

16  Drams 

16  Ounces 

28  Pounds 

4  Quarters 


make  1  Ounce 
1  Pound 
1  Quarter 
1  Hundred  Weight 


marked  dr 
oz 


20  Hundred  Weight  1  Ton 
dr 


lb 
qr 

cwt 
ton 


16  = 

256  = 

7168  = 

28672  = 


oz 
1 

16  = 
448  = 
1792 


lb 

1  qr 
28  =  1. 
112  =  4     = 


cwt 

1 


573440  =  35840  —  2240  =  80  =  20  =  1 

By  this  weight  are  weighed  all  things  of  a  coarse  or  drossy 
nature,  as  Corn,  Bread,  Butter,  Cheese,  Flesh,  Grocery 
Wares,  and  some  Liquids  j  also  all  Metals,  except  Silver  and 
Gold. 

oz  dwt  gr 
Abfc,  that  1Z6  Avoirdupois  =  14  11   15i  Troy. 
loz         -         -     =    0  18     6i 
Idr        -         -     =    0     1     3| 

Hence  it  appears  that  the  pound  Avoirdupois  contains  6999^ 
grains,  and  the  pound  Troy  5760  ;  the  former  of  which  aug- 
mented by  half  a  grain  becomes  7000,  and  its  ratio  to  the  latter 
is  therefore  very  nearly  as  760  to  576,  that  is,  as  175  to  144  ; 
consequently  144  pounds  Avoirdupois  are  very  nearly  equal  to 
175  pounds  Troy  :  and  hence  we  infer  that  the  ounce  Avoirdu- 
pois is  to  the  ounce  Troy  as  175  to  192. 


LONG  MEASURE. 

3  Barley-corns  make 

1  Inch 

•    In 

12  Inches 

1  Foot 

-    Ft 

3  Feet          ^ 

1  Yard        - 

-     Yd 

6  Feet 

1  Fathom    - 

'    Fth 

5  Yards  and  a  half 

1  Pole  or  Rod 

-    PI 

40  Poles 

1  Furlong   - 

-     Fur 

8  Furlongs  - 

1  Mile 

-    Milt 

3  Miles 

1  League    - 

-     Lea 

69yV  Miles  nearly  - 

1  Degree    - 

-     Deg  or  ^. 

weights  one  ounce,  and  12  ounces  one  pound.  But  in  latter  times,  it 
was  thought  sufficient  to  divide  the  same  pennyweight  into  24  equal 
.parts,  still  called  grains,  being  the  least  weight  now  in  common  use  ; 
»nd  from  thence  the  rest  are  computed,  as  in  the  Tables  above. 

In 


TABLES  OF  MEASURES.  27 


In              Ft 

12=         1                Yd          . 

36  =         3    =         1              PI 

198  =       16^  =         6|  =    .    1 

Fur, 

920  =     660    =     220    =     40  = 

1 

MUi 

i360  =  6280    =1760    =  320  = 

8  = 

1 

CLOTH  MEASURE. 

2  Inches  and  a  quarter  make  I  Nail  ■         -  -  NX 
4  Nails         -         -         -  1  Quarter  of  a  Yard  (^r 

3  Quarters  -         -         -  1  Ell  Flemish  -  E  F 

4  Quarters  -         -         -  1  Yard          -  -  Yd 
6  Quarters   -         -         -  1  Ell  English  -  E  E 
4  Quarters  1|  Inch       -  1  Ell  Scotch  -  ES 


SQUARE  MEASURE. 

144     Square  Inches  make  1  Sq  Foot  -  Ft 

9     Square  Feet      -  1  Sq  Yard  <-  Yd 

30i  Square  Yards    -  1  Sq  Pole  -  PoU 

40     Square  Poles      -  1  Rood     -  -  Rd 

4     Roods        -         -  1  Acre     -  -  Acr 

Sq  Inch  Sq  Ft 

144  =  1  SqYd 

1296=  9    =  1  SqPl 

39204  =  2721  =  301  =  1          Rd 

1568160  =  10890    =  1210     =  40  =      1         Acr 

6272640  =  43560     =  4840    =  160  =     4  =      1 

By  this  measure,  Land,  and  Husbandmen  and  Gardeners' 
work  are  measured  ;  also  Artificers'  work,  such  as  Board, 
Glass,  Paveor^ents,  Plastering,  Wainscoting,  Tiling,  Flooring, 
and  every  dimension  of  length  and  breadth  only. 

When  three  dimensions  are  concerned,  namely,  length, 
breadth,  and  depth  or  thickness,  it  is  called  cubic  or  solid 
measure,  which  is  used  to  measure  Timber,  Stone,  &c. 

The  cubic  or  solid  Foot,  which  is  12  inches  in  length  and 
breadth  and  thickness,  contains  1728  cubic  or  solid  inches., 
and  27  solid  fe$jt  make  one  solid  yard. 


DRY 


2« 


2  Pints  make 

2  Quarts 

2  Pottles       - 

2  Gallons      - 

4  P'ecks 

8  Bushels 

5  Quarters    - 
2  Weys 


'^^■--~-.-  i 

ARITHMETIC.     >5- 

DRY,  OR  CORN  MEASURE. 

1  Quart  -  -  qt 

1  Pottle  -  -  Pot 

1  Gallon  -  -  Qcd 

1  Peck  -  -  Pec 

1  Bushel  -  -  Bu 

1  Quarter  -  -  Qr 

1  Wey,  Load,  or  Ton  Wey 

1  Last     -  -  -  Last 


Gal 
1 


Pts 

8  = 

16  =  2  = 

64  =  8  = 

612  =  64  = 

2660  =  320  = 

6120=  640  = 


Pec 

1         Bu 

4=1  Qr 

32  =     8  =       1         Wey 
160  =  40  =       5  =       1         Last 

320  =  80  =  10  =       2  =       1 


By  this  are  measured  all  dry  wares,  as,  Corn,  Seeds,  Roots, 
Fruit,  Salt,  Coals,  Sand,  Oysters,  &c. 

The  standard  Gallon  dry  measure  contains  26 8|- cubic  or 
solid  inches,  and  the  Corn  or  Winchester  bushel  2150|  cubic 
inches  ;  for  the  dimensions  of  the  Winchester  bushel,  by  the 
Statute,  are  8  inches  deep,  and  18^  inches  wide  or  in  diameter, 
but  the  Coal  bushel  must  be  19^  inches  in  diameter  ;  and  36 
bushels,  heaped  up,  make  a  London  chaldron  of  coals,  the 
weight  of  which  is  31561b  Avoirdupois. 

ALE  AND  BEER  MEASURE.        ^,       *^^ ^ 


A  Hi 


2  Pints  make 

- 

1  Quart 

- 

Qt 

4  Quarts 

- 

1   Gallon 

- 

Gal 

36  Gallons      - 

- 

4  Barrel 

- 

Bar 

1  Barrel  and  a 

half 

1  Hogshead 

Hhd 

2  Barrels     - 

. 

1  Puncheon 

Pun 

2  Hogsheads 

- 

1  Butt 

- 

Butt 

2  Butts 

- 

1  Tun 

- 

Tun 

Pts           qt 

2=         1 

Gal 

8  =         4  = 

1 

Bar 

288  =     144  = 

36  = 

1 

ma 

432  =     216  = 

54  = 

--     H  = 

1 

Butt 

864  =     432  = 

108  = 

=      3    = 

2  = 

1 

JVbfe,  The  Ale  Gallon  contains  282  cubic  or  solid  inches. 


WINE 


TABLES  OF  MEASURES  AND  TIME. 


WINE  MEASURE. 


2  Pints  make 

4  %iarts 

42  Gallons 

63  Gallons  or   1^  Tierces 

2  Tierces 

2  Hogsheads 

2  Pipes  or  4  Hhds 


1  Q^art        -  Q^ 

1  Gallon        -  G<d 

1  Tierce       -  Tier 

1  Hogshead  -  Hhd 

1  Puncheon  -  Pirn 

1  Pipe  or  Butt  Pi 

I  Tun      -     -  Tun 


Pis 

2  = 

8  = 

336  = 

604  = 

672  = 

10Q8  = 


qt 

1 

4  = 
168  = 

262  = 
336  = 


Gal 

1       Tier 
42  =     1         Hhd      ^ 
63  =     U  =     1        Pun 
84  =     2"  =     U  =  1 
604  =     126  =     3    =     2^  ='n 


1      Tmi 
2016  =     1008  =     252  =     6    =     4    =  3""  =  2  =    1 


JVo/e,  By  this  are  measured  all  Wines,  Spirits,  Strong- 
waters,  Cyder,  Mead,  Perry,  Vinegar,  Oil,  Honey,  &c. 

The  Wine  Gallon  cantains  231  cubic  or  solid  inches.  And 
it  is  remarkable,  that  the  Wine  and  Ale  Gallons  have  the  same 
proportion  to  each  other,  as  the  Troy  and  Avoirdupois  Pounds 
have  J  that  is,  as  one  Pound  Tray  is  to  one  Pound  Avoirdu- 
pois, so  is  one  Wine  Gallon  to  one  Ale  Gallon. 

OF  TIME. 


60  Seconds  or  60"  make 

60  Minutes 

24  Hours  -         -         -         - 

7  Days    -         -         -         - 

4  Weeks 

13  Months  1  Day  6  Hours,  } 
or  365  Days  6  Hours  ^ 


1  Minute 

1  Hour 

1  Day 

1  Week 

1  Month 


M  or  ' 

Hr 

Day 

JVk 

Mo 


1  Julian  Year  Yr 


Sec 

60  = 

3600  = 

36400  = 

604800  = 

2419200  = 

31657600  = 


Mill 

1 

60  = 

1440  = 

10080  = 

40320  = 

525960  = 


Hr 

1 

24  = 

168  = 

672  = 


8766  =  366A  =r 


Day 

1        Wk 

'7    =^  1      Mq 
28  =4=1 


1  Year 


Or 


30  ARITHMETIC. 

Wk  Da  Hr     Mo  Da  Hr, 
Or  52     1     6=13    1     6=1  Julian  Year 
Da   Hr    M  Sec 
But  365    5    48    48  =  1  Solar  Year. 


RULES  FOR  REDUCTION. 

I.  When  the  Numhers  are  to  he  reduced  from  a  Higher  Deno- 
mination to  a  Lower  : 

Multiply  the  number  in  the  highest  denomination  by  as 
many  as  of  the  next  lower  make  an  integer,  or  1,  in  that 
higher  ;  to  this  product  add  the  number,  if  any,  which  was 
in  this  lower  denomination  before,  and  set  down  the  amount. 

Reduce  this  amount  in  like  manner,  by  multiplying  it  by 
as  many  as  ©f  the  next  lower,  make  an  integer  of  this,  taking 
in  the  odd  parts!  of  this  lower,  as  before.  And  so  proceed 
through  all  the  denominations  to  the  lowest ;  so  shall  the 
number  last  found  be  the  value  of  all  the  numbers  which 
were  in  the  higher  denominations,  taken  together.* 

EXAMPLE. 

1.  In  1234Z  15s  Id^  how  many  farthings  ? 
I       s     d 
1234    15   7 
20 


24695  Shillings 
12 


296347  Pence 
4 


Answer  11 85388  Farthings. 


♦  The  reason  of  this  rule  is  very  evident ;  for  pounds  are  brought 
into  shillings  by  multiplying  them  by  20 ;  shillings  into  pence,  by  mul- 
tiplying them  by  12 ;  and  pence  into  farthings,  by  multiplying  by  4  ; 
and  the  reverse  of  this  rule  by  Division— And  the  same,  it  is  evident, 
vrill  be  true  in  the  reduction  of  numbers  consisting  of  any  denomina- 
tions "whatever. 

n.  When 


RULES  FOR  REDUCTION.  31 


II.  When  the  Numbers  are  to  be  reduced  from  a  Lower  Denomi- 
nation to  a  Higher : 

DivipE  the  given  number  by  as  many  as  of  that  denomina- 
tion make  1  of  the  next  higher,  and  set  down  what  remains, 
as  well  as  the  quotient. 

Divide  the  quotient  by  as  many  as  of  this  denomination 
make  1  of  the  next  higher  ;  setting  down  the  new  quotient, 
and  remainder,  as  before. 

Proceed  in  the  same  manner  through  all  the  denomina- 
tions, to  the  highest ;  and  the  quotient  last  found,  together 
with  the  several  remainders,  if  any,  will  be  of  the  same  value 
as  the  first  number  proposed. 


EXAMPLES. 

2.  Reduce    1186388  farthings  into    pounds,  shillings,  and 
pence. 

4)  1186388 

12)    2.96347  ci 


2,0)      2469,6  s—ld 
Answer   1234/  15s  Id 


3.  Reduce  24/  to  farthings.  Ans.  23040. 

4.  Reduce  337687  farthings  to  pounds,  &c. 

Ans.  351/135  Of. 

5.  How  many  farthings  are  in  36  guineas  ?  Ans.  36288. 

6.  In  36288  farthings  how  many  guineas  ?  Ans.  36. 

7.  In  59  lb  13  dwts  5  gr  how  many  grains  ?      Ans.  340157. 

8.  In  8012131  grains  how  many  pounds,  &c.  ? 

Ans.  13901b  11  oz  18  dwt  19  gr. 

9.  In  35  ton  17  cwt  1  qr  23  lb  7  oz  13  dr  how  many  drams  ? 

Ans.  20571006. 

10.  How  many  barley-corns  will  reach  round  the  earth, 
supposing  it,  according  to  the  best  calculations,  to  be  25000 
miles  ?  Ans.  4762000000. 

1 1 .  How  many  seconds  are  in  a  solar  year,  or  366  days 
5  hrs  48  min  48  sec  ?  Ans.  31556928. 

12.  In  a  lunar  month,  or  29  ds  12  hrs  44  min  3  sec,  how 
many  second*  ?  •  Ans.  2551443. 

COM- 


32  ARITHMETIC. 


COMPOUND  ADDITION. 

Compound  Addition  shows  how  to  add  or  collect  several 
nambers  of  different  denominations  into  one  aafti. 

Rule. — Place  the  numbers  so,  that  those  of  the  same  de- 
nomination may  stand  directly  under  each  other,  and  draw  a 
line  below  them.  Add  up  the  figures  in  the  lowest  denomi- 
nation, and  find,  by  Reduction,  how  many  units,  or  ones,  of 
the  next  higher  denomination  are  contained  in  their  sum. — 
Set  down  the  remainder  below  its  proper  column,  and  carry 
those  units  or  ones  to  the  next  denomination,  which  add  up 
in  the  same  manner  as  before. — Proceed  thus  through  all  th« 
denominations,  to  the  highest;  whose  sum,  together  with  the 
several  remainders,  will  give  the  answer  sought. 

The  method  of  proof  is  the  same  as  in  Simple  Addition. 

EXAMPLES  OF  MONEY. 

1.  2.  3.  4. 

I      s      d  I       s      d  I      s      d  I      s      d 

"        7   13     3         14     7     5  15  17  10         53  14     8 


3  5 

lOi 

8*19 

2i 

3 

14 

6 

5  10 

n 

6  18 

7 

7  8 

H 

23 

6 

n 

93  11 

6 

0  2 

^ 

21  2 

9 

14 

9 

41 

7  6 

0 

4  0 

3 

7  16 

SI 

15 

6 

4 

13  2 

5 

17  15 

H 

0  4 

3 

6 

12 

9f 

0  18 

7 

39  15 

n 

32  2 

^ 

39  15 

n 

6. 

6. 

7. 

8. 

I      s 

d 

I      s 

d 

I 

s 

d 

I      s 

d 

14  0 

71" 

37  15 

8 

61 

3^ 

H 

472  15 

3 

8  15 

3 

14  12 

n 

7 

16 

8 

9  2 

n 

62  4 

7 

17  14 

9 

29 

13 

lOf 

27  12 

H 

4  17 

8 

23  10 

n 

12 

16 

'2 

370  16 

H 

23  0 

4f 

8  6 

0 

0 

7 

H 

13  7 

4 

6  6 

7 

14  0 

5i 

24 

13 

0 

6  10 

H 

91  0 

101 

64  2 

^ 

5 

0 

lOf 

30  0 

111 

* 

EXAM- 


COMPOUND  ADDITION. 


33 


Exam.  9.  A  nobleman,  going  out  of  town,  is  informed  by 
his  steward  that  his  butcher's  bill  comes  to  197/  13s  l^d;  his 
baker's  to  69Z  5s  2^d ;  his  brewer's  to  83/ ;  his  wine  mer- 
chant's to  103/  13s;  to  his  corn-chandler  is  due  76/  3d ;  to 
his  tallow-chandler  and  cheesemonger,  27/  15s  \\\d;  and 
to  his  tailor  55/  3s  6|(/ ;  also  for  rent,  servants'  wages,  and 
other  charges,  127/ 3s:  Now,  supposing  he  would  take  100/ 
with  him  to  defray  his  charges  on  the  road,  for  what  sum 
must  he  send  to  his  banker  ?  Ans.  830/  14s  ^d. 

10.  The  strength  of  a  regiment  of  foot,  of  10  companies, 
and  the  amount  of  their  subsistence*,  for  a  month  of  30  days, 
according  to  the  annexed  Table,  are  required  ? 


Numb. 

Rank. 

Subsistence  for 

a  Month. 

/     * 

d 

Colonel 

27     0 

0 

Lieutenant  Colonel 

19   10 

0 

Major 

17     6 

0 

Captains 

78   15 

0 

U 

Lieutenants 

67  16 

0 

Ensigns 

40  10 

0 

Chaplain 

7  10 

0 

Adjutant 

4  10 

0 

Quarter- Master 

6     5 

0 

Surgeon 

4   10 

0 

Surgeon's  Mate 

4   10 

0 

30 

Serjeants 

45     0 

0 

30 

Corporals 

to    0 

0 

20 

Drummers 

20     0 

0 

2 

Fifers 

2     0 

0 

390 

Private  Men 

292  10 

0 

607 

Total 

656   10 

0 

*  Subsistence  Money,  is  the  nioney  paid  to  the  soldiers  weekly, 
•which  is  short  of  their  full  pay,  because  their  clothes,  accoutrements, 
&c.  are  to  be  accounted  for.  It  is  likewise  the  money  advanced  to 
officers  till  their  accounts  are  made  up,  which  is  commonly  once 
a  year,  when  they  are  paid  their  arrears.  The  following  Table  shows 
the  full  pay  and  subsistence  of  each  rank  on  the  Eng^Ush  establish- 
ment. 


Vol.  L 


BAILY 


34 


ARITHMETIC. 


O 


O  CO  CD  CO  Q 

05  CO  'it  CO  t^ 

->      ^      ^      o  o 


CO  O 

O         '-<         CO         ©<  CD 
T-H         ^         O         O  O 


O  O         00  O  O  C 
lO  '^         u:3  ^  O  »0 

o  o      o  o  o  o 


CO  O 
r}<  CO 

o  o 


CO  CO  O 
I   (M  l>  10 

OOP 


O 


»o 


^ 


o  o 
o      o  o 


-CO 


O  CO  CO  CO  CO  O  CO  CO  O  O 

'—         G^»-i         COt-*  ^[rf         COlOiVO 

^         ^         ^         O  O O        O         OOP 


■P 

p 

OPP  P 

p 

p 

1 

3 

CO 

;;^ 

CO  Tl*  CO  -^ 

1 1  = 

1  r  1 

(24 

m 

^^ 

-H    ^    O    P 

p 

p 

^ 

o 

CO 

CO  p  »^  o< 

CO 

p  ■ 

S 
03 

1> 

CO 

Oi  CO  ©<  co» 

P  P  p  o 

1  r 

p 

1  l'^  1 

p 

<'C 


CO       ^       '-I 


G^ 


lO 


T-*  P 


P     P     P     P     P 
rf    Tt    P    l>    lO 

Qj  ^  ^  o  o 


^  C»  CO 

J  O)  u:)  lo 

p  p  p 


CO  P 
I   rf  lO 


P5 


CO 


p  p  p  p, 
p  i^  CO  io 
P  p  p  p 


iC  Ci     j   vo 

OP  P 


I  I  I 


©<     I    CO     I    O) 

'^      ^      p 


p  p      p 

O  «0      I     -H 


j>  p  o 
^  C5  Ol 

p  p  p 


I"  I 


P  P  Q  P 
lO  P  O)  CO 
P  P  P  P 


P 


P  "^         P~P~ 

lO  »-^    I  »o  CO 


I  I 


J    G>l     j    CO  CO 

p      p  p 


I"" 


oj  <u  a>  :^ 
a   a  -^ 


"SScg 


>  *^ J  s  ^ 


(^j  Q  O  ^  G^ 


4j      .    C 


O     G  T3' 

■;j  LJ  < 


P-i  Qrc/3  CQ  <!  >  CO 


<s 

*^f, 

^?. 

^^ 

1^ 

^2 

I.    4> 

O^-o 

h  u 

O  J3 

^   S 

la 

"3*^3 

11 

et  to 
C.S 

0 


EXAMPLES 


COMPOUND  ADDITION. 


35 


EXAMPLES  OF  WEIGHTS,  MEASURES,  &c. 


TROY  WEIGHT. 

I 

LPOTHE 

;CAR 

lES'  1 

tVEIGHT. 

1. 

2. 

3. 

4. 

lb  oz  dwt 

oz  dwt  gr 

lb 

oz   dr 

sc 

OZ 

dr  sc  gr 

17     3   13 

37     9     3 

3 

6     7 

2 

3 

5     1    17 

7     9     4 

9     6     3 

13 

7     3 

0 

7 

3     2     5 

0  10     7 

8   12   12 

19 

10     6 

2 

16 

7     0  12 

9     5     0 

17     7     8 

0 

9     1 

2 

7 

3     2     9 

176     2   17 

6     9     0 

36 

3     6 

0 

4 

1     2  18 

23  11    12 

3     0  19 

5 

8     6 

1 

36 

4     1   14 

AVOIRDUPOIS  WEIGHT. 

LONG  MEASURE. 

5. 

6. 

7. 

8. 

lb  oz  dr 

cwt  qr  lb. 

mis  fur  pis 

j^ds  feetinc 

-  17   10  13 

15     2   15 

29     3  14 

127     1     5 

5  14     8 

6     3  24 

19     6  29 

12     2     9 

12     9   18 

9     1    14 

7     0  24 

10     0  10 

27     1     6 

9     1    17 

9     1  37 

54      1    11 

0     4     0 

10     2     6 

7     0     3 

5     2     7 

6  14  10 

3     0     3 

4     5     9 

23     0     5 

CLOTH  MEASURE. 

LAND  MEASURE. 

9. 

10. 

11. 

12. 

yds  qr  nls 

el  en  qrs  nls 

ac   ro   p 

ac     ro  p 

26     3     1 

270     I     0 

225     3  37 

19     0   16 

13     1     2 

57     4     3 

16      1   25 

270     3  29 

9     1     2 

18     1     2 

7     2   18 

6     3  13 

217     0     3 

0     3     2 

4     2     9 

23     0  34 

9     1     0 

10     1     0 

42      1    19 

7     2,16 

a5    3    1 

4     4     1 

7     0     6 

75     0  23 

WINE  MEASURE. 

ALE  AND  BEER  MEASURE. 

13. 

14. 

15. 

16. 

t  hds  gal 

hds  gal  pts 

hds  gal  pts 

hds  gal  pts 

13     3  16 

15  61     5 

17  37     3 

29  43     5 

8     1  37 

17   14   13 

9   10   15 

12   19     % 

14     1   20 

29  23     7 

3     6     2 

14   16     6 

25     0    12 

3   15     1 

5   14     0 

6     8      1 

3     1     9 

16     8     0 

12     9     6 

57   13     4 

72     3  21 

4  36     6 

8  42     4 

5     6     0 

COM- 


i&  ARITHMETIC, 


COMPOUND  SUBTRACTION; 


Compound  Subtraction  shows  how  to  find  the  difference 
between  any  two  numbers  of  different  denominations.  To 
perform  which,  observe  the  following  Rule  : 

*  Place  the  less  number  below  the  greater,  so  that  the 
parts  of  the  same  denomination  may  stand  directly  under 
each  other  ;  and  draw  a  hne  below  them. — Begin  at  the 
right-hand,  and  subtract  each  number  or  part  in  the  lower 
line,  from  the  one  just  above  it,  and  set  the  remainder 
straight  below  it.  But  if  any  number  in  the  lower  line  be 
greater  than  that  above  it,  add  as  many  to  the  upper  number 
as  make  1  of  the  next  higher  denomination  ;  then  take  the 
lower  number  from  the  upper  one  thus  increased,  and  set 
down  the  remainder.  Carry  the  unit  borrowed  to  the  next 
number  in  the  lower  line  ;  after  which  subtract  this  number 
from  the  one  above  it,  as  before  ;  and  so  proceed  till  the  whole 
is  finished.  Then  the  several  remainders,  taken  together, 
will  be  the  whole  difference  sought. 

The  method  of  proof  is  the  same  as  in  Simple  Subtraction. 


EXAMPLES  OF  MONEY. 


1. 

I      s 
From  79  17 
Take  35  12 

d 

2. 

I       s 

103  3 

71  12 

d 

n 

I 

81 
29 

5. 
s 

10 
13 

d 
11 

I 

254 
37 

4. 

s  d 
12  0 
9  4| 

Rem.  44  5 

"^1 

31  10 
103  3 

8| 

Proof.  79  17 

6.  What  is  the  difference  between  73Z  b^d  and  19/135  lOd  ? 

Ans.  53/  65  l\d. 


*  The  reasoTi  of  this  Rule  will  easily  appear  from  what  has  been 
said  in  Simple  Subtraction  ;  for  the  borrowing  depends  on  the  same 
principle,  and  is  only  different  as  the  numbers  to  be  subtracted  are  of 
different  denominations* 

Ex.  6 


COMPOUND  SUBTRACTION. 


^.7 


Ei.  6.  A  lends  to  B  100/,  how  much  is  B  in  debt  after  A 
has  taken  goods  of  him  to  the  amount  of  73/  I2s  4^d? 

Ans.  26/  75  7i(?. 

7.  Suppose  that  my  rent  for  half  a  year  is  20/  125,  and  that 
I  have  laid  out  for  the  land-tax  14»  6d,  and  for  several  repairs 
1/  3s  3i</,  what  have  1  to  pay  of  my  half-year's  rent  ? 

Ans.  Vcl  145  2Srf. 

8.  A  trader  failing,  owes  to  A  35/  75  6c/,  to  B  91/  13s  ^d, 
to  C  53/  7irf,  to  D  87/  55,  and  to  E  111/  35  5^d.  When  this 
happened,  he  had  by  him  in  cash  23/  75  5c/,  in  wares  53/  1  Is 
lO^e/,  in  household  furniture  63/  175  7|</,  and  in  recoverable 
book-debts  25/  7s  5d.  What  will  his  creditors  lose  by  him, 
suppose  these  things  delivered  to  them  ?         Ans.  212/  5s  S^d. 


EXAMPLES  OF  WEIGHTS,  MEASURES,  &c. 

TROY  WEIGHT.  APOTHECARIES'  WEIGHT. 


1. 


2. 


lb  oz  dwt  gr 
From  9    2    12   10 
Take  5    4     6   17 

lb  oz  dwt  gr 
7  10     4  17 
3     7   16   12 

lb  oz  dr  scr  gr 
73     4    7    0     14 

29     5    3    4     19 

Rem. 

Proof 

AVOIRDUPOIS  WEIGHT. 
4.  5. 

c  qrs  lb  lb  oz  dr 
From  6  0  17  71  5  9 
Take    2    3    10         17     9   18 


LONG  MEASURE. 

6.  7. 

m  fu  pi  yd  ft  in 
14  3  17  96  0  4 
7     6   11  72     2     9 


Rem. 

ac 

17 
16 

Proof 

CLOTH  MEASURE. 
8.                   9. 
yd   qr  nl       yd  qr   nl 
From  17     2     1         9    0    2 
Take     9     0     2         7    2     1 

LAND  MEASURE. 

10.                      11. 
ro    p         ac   ro  p 

1  14         57     1    16 

2  8         22     3  29 

Rem. 

Proof 

^VINfi 


38  ARITHMETIC. 

WINE  MEASURE.  ALE  AND  BEER  MEASURE. 


12. 

13. 

14. 

15. 

t   hdgal 

hd  gal  pt 

hd  gal  pt 

hd   gal   pt 

From   17    2  23 

5     0     4 

14    29     3 

71     16     6 

Take     9    1  36 

2   12     6 

9  35     7 

19      7     1 

Rem. 
Proof 


DRY  MEASURE. 

TIME. 

From 
Take 

la 

9 
6 

16.                    17. 

qr  bu  bu  gal  pt 
4  7  13  7  1 
3     6         9      2     7 

18.                       19. 
mo  we  da        ds   hrs   min 
71      2     5        114     17     26 
17      1     6         72      10     37 

Rem. 
Proof 


20.  The  line  of  defence  in  a  certain  polygon  being  236 
yards,  and  that  part  of  it  which  is  terminated  by  the  curtain 
and  shoulder  being  146  yards  1  foot  4  inches  ;  what  then  was 
the  length  of  the  face  of  the  bastion  ?       Ans.  89  yards  1  ft  8  in. 


COMPOUND  MULTIPLICATION. 


Compound  Multiplication  shows  how  to  find  the  amount 
of  any  given  number  of  different  denominations  repeated  a 
certain  proposed  number  of  times  j  which  is  performed  by 
the  following  rule. 

Set  the  multiplier  '  under  the  lowest  number  of  the 
multiplicand,  and  draw  a  line  below  it. — Multiply  the  num- 
ber in  the  lowest  denomination  by  the  multiplier,  and  find 
how  many  units  of  the  next  higher  denomination  are  con- 
tained in  the  product,  setting  down  what  remains. — In  like 
manner,  multiply  the. number  in  the  next  denomination,  and 
to  the  product  carry  or  add  the  units,  before  found,  and  find 
how  many  units  of  the  next  higher  denomination  are  in  this 

amount. 


COMPOUND  MULTIPLICATION.  39 

amount,  which  carry  in  like  manner  to  the  next  product^ 
settins;  down  the  overplus. — Proceed  thus  to  the  highest  de- 
nonauation  proposed  :  so  shall  the  last  product,  with  the  se- 
Teral  remainders,  taken  as  one  compound  number,  be  the 
whole  amount  required. — The  method  of  Proof,  and  the 
reason  of  the  Rule,  are  the  same  as  in  Simple  Multiplicatioo. 


EXAMPLES  OF  MONEY. 

To  find  the  amount  of  81b  of  Tea,  at  5s  8^d  per  lb. 
s     d 
6     8| 
8 


£2    5,8  Answ^. 

/     s  d 

2.  4  lb  of  Tea,  at  7s  8d  per  lb.  Ans.     1   10  8 

3.  6  lb  of  Butter,  at  9^^  per  lb.  Ans.    0     4  9 

4.  7  lb  of  Tobacco,  at  Is  S^d  per  lb.         Ans.    0  11  lli^ 

5.  9  Stone  of  Beef,  at  2s  1^  per  st.  Ans.     1     1  0 

6.  10  cwt  of  Cheese,  at  n  ifs  lOd  per  cwt.  Ans.  28   1 8  4 

7.  12  cwt  of  Sugar,  at  31  Is  Ad  per  cwt.         Ans.  40     8  0 

CONTRACTIONS. 

L  If  the  multiplier  exceed  12,  multiply  successively  by  its 
component  parts,  instead  of  the  whole  number  at  once. 

EXAMPLES. 


IS  cwt  of  Cheese,  at  17*  6d  per  cwt. 
/     s    d 
0  17     6 
3 


12     6 
6 


13     2     6  Answer. 


2.  20  cwt  of  Hops,  at  Al  Is  2d  per  cwt.       Ans.  87     3    4 

3.  24  tons  of  Hay,  at  31  Is  6rfper  ton.         Ans.  .81     0     0 

4.  45  ells  of  Cloth,  at  1«  Qd  per  ell.  Ans.    3    7    6 

Ex.  5. 


•W 


4©        ^  ARITHMETIC. 

/  *    d 

Ex.6.  63  gallons  of  Oil,  at  2s  3d  per  gall.        Ans.      7  19 

6.  70  barrels  of  Ale,  at  11  45  per  barrel      Ans.    84  0     0 

7.  84  quarters  of  Oats,  at  1/  12s  8d  per  qr.  Ans.  137  4     0 

8.  96  quarters  of  Barley,  at  1/  3s  4d  per  qr.  Ans.  112  0     0 

9.  120  days'  Wages,  at  5s  9d  per  day.         Ans.    34  10     0 

10.  144  reams  of  Paper,  at  13s  4d  per  ream.  Ans.    96     0     0 

11.  If  the  multiplier  cannot  be  exactly  produced  by  the 
multiplication  of  simple  numbers,  take  the  nearest  number  to 
it,  either  greater  or  less,  which  can  be  so  produced,  and  mul- 
tiply by  its  parts,  as  before. — Then  multiply  the  given  mul- 
tiplicand by  the  difference  between  this  assumed  number  and 
the  multiplier,  and  add  the  product  to  that  before  found,  when 
the  assumed  number  is  less  than  the  multiplier,  but  subtract 
the  same  when  it  is  greater. 

%       •  EXAMPLES. 

1.  26  yards  of  Cloth,  at  3s  O^d  per  yard. 
I     s    d 
0     3    0| 
6 


0  16     3f 
5 

3  16     6^ 
3     Of 


£  3  19     7i  Answer. 


I  s  d 

2. 29quartersofCorn,at2/5s3ic?perqr.    Ans.    65  12  loi 

3.  63  loads  of  Hay,  at  3^  15s  2d  per  load-     Ans.  199  3  10* 

4.  79  bushels ofWheat. at  lis  5|rf per  bush.  Ans.    45  6  10^ 

5.  97  casks  of  Beer,  at  12s  2d  per  cask.       Ans.    69  0  2 

6.  114stoneofMeat,at  15s3f(/perst0ne.    Ans.    87  6  71. 

EXAMPLES  OF  WEIGHTS  AND  MEASURES. 

I.  2.  3. 

lb  oz  dwt  gr     lb  oz  dr  sc  gr    cwt  qr  lb  oz 

28   7   14  10      2  6  3  2  10    29  2  16  14 

5  8  12 


^- 


COMPOUND  DIVISION.  41 


4. 

mis    fu   pis  yds 

22     5     29     6 

4 

5. 
yds       qrs    na 
126        3       1 
7 

mo 
172 

6. 
ac       ro        po 
28       3        27 
9 

7. 
tuns  hhd  gal  pts 
20     2     26     2 
3 

8. 
we  qr   bu   pe 
24     2     5       3 
6 

9. 
we   da   ho  min 
3     6     16     49 
10 

COMPOUND  DIVISION. 


Compound  Division  teaches  how  to  divide  a  number  of 
several  denominations  by  any  given  number,  or  into  any  num> 
ber  of  equal  parts  ;  as  follows  : 

Place  the  divisor  on  the  left  of  the  dividend,  as  in  Simple 
Division. — Begin  at  the  left-hand,  and  divide  the  number  of 
the  highest  denomination  by  the  divisor,  setting  down  the 
quotient  in  its  proper  place. — If  there  be  any  remainder  after 
this  division,  reduce  it  to  the  next  lower  denomination,  which 
add  to  the  number,  if  any,  belonging  to  that  denomination,  and 
divide  the  sum  by  the  divisor. — Set  down  again  this  quotient, 
reduce  its  remainder  to  the  next  lower  denomination  again, 
and  so  on  through  all  the  denominations  to  the  last. 


EXAMPLES  OF  MONEY. 

1.  Divide  237/  8«  Gd  by  2. 
I      s     d 
9)  237     8     6 


ig  118  14     3  the  Quotient. 

2.  Divide 


Vol.  I. 


4t  ARITHMETIC. 

I  s  d  I        s  d 

2.  Divide     432  12  If  by    3.  Ana.  144     4  Oi 

3.  Divide     607  3  5    by    4.  Ans.  126  15  lOi 

4.  Divide     632  7  6^  by    6.  Ans.  126     9  6 
6.  Divide     690  14  3i  by    6.  Ans.  ,115     2  4i 

6.  Divide     705  10  2    by    7.  Ans.  100  15  8f 

7.  Divide  760  5  6  by  8.  Ans.  95  0  8^ 
S.Divide  761  5  7f  by  #.  Ans.  84  11  ^ 
9.  Divide     829  17  10    by  10.  Ans.  82  19  9| 

.10.  Divide     937  8  8fbyll.  Ans.  86     4  5 

11.  Divide  H45  11  4i  by  12.  Ans.  95     9  31 


CONTRACTIONS. 

I.  If  the  divisor  exceed  12,  find  what  simple  numbers, 
multiplied  together,  will  produce  it,  and  divide  by  them  sepa- 
rately,  as  in  Simple  Division,  as  below. 


EXAMPLES. 

1.  What  is  Cheese  per  cwt,  if  16  cwt  cost  25f  14s  ^d? 
I      s      d 
4)25     14     8 


4)    6       8     8 


£   1     12    2  the  Answer. 


2.  If  20  cwt  of  Tobacco  come  to  } 
1501  6s  M,  what  is  that  per  cwt  ?  ^ 


I    s    d 
Ans.  7  10  4 


3.  Divide   98/  8s  by  36.  Ans.  2  14  8 

4.  Divide  ^1/  13s  lOd  by  66.  Ans.       15  7i 

5.  Divide  44/  4s  by  96.  Ans.  0     9  2^ 
^.  At  31/  10s  per  cwt,  how  much  per  lb  ?  Ans.  0    5  7^ 

II.  If  the  divisor  cannot  be  produced  by  the  multiplication 
of  small  numbers,  divide  by  the  whole  divisor  at  once,  after 
the  manner  of  Long  Division,  as  follows. 


?XAM- 


COMPOUND  DIVISION.  43 


EXAMPLES. 


1.  Divide  59/ 6*  3ft?  by  19. 
I    s  d  I  s   d 

19)  59  6  3f         (  3  2  ai  Ans. 
57 

2 
20 

46(2 
38 

8 

12 

99(5 
95 

4 
4 

19(1 


/ 

9 

d 

Ans. 

0 

13 

Hi 

Ads. 

2 

18 

3 

Ans. 

6 

11 

10 

Ans. 

0 

19 

5. 

I      s      d 

2.  Divide     39   14     5i  by  67. 

3.  Divide   125     4     9    by  43|. 

4.  Divide  542     7  10    by  97. 

5.  Divide  123  11     2i  by  127. 


EXAMPLES  OF  WEIGHTS  AND  MEASURES. 

1.  Divide  17  lb  9  oz  0  dwts  2  gr  by  7. 

Ans.  2  lb  6  oz  8  dvirts  14gr. 

2.  Divide  17  lb  5  oz  2  dr  1  scr  4  gr  by  12. 

Ans.  1  lb  5  oz  3  dr  1  scr  12  gr. 

3.  Divide  178  cwt  3  qrs  14  lb  by  53.  Ans.  3  cwt  1  qr  14  lb. 

4.  Divide  144  mi  4  fur  2  po  1  yd  2  ft  0  in  by  39. 

Ans.  3  mi  5  fur  26  po  0  yds  2  ft  8  in. 

5.  Divide  534  yds  2  qrs  2  na  by  47.     Ans.  1 1  yds  1  qr  2  na. 

6.  Divide  7 1  ac  1  ro  33  po  by  51.  Ans.  1  ac  2  ro  3  po. 

7.  Divide  7  tu  0  hhds  47  gal  7  pi  by  65.      Ans.  27  gal  7  pi. 

8.  Divide  387  la  9  qr  by  72.  Ans.  5  la  3  qrs  7  bu. 

9.  Divide  206  mo  4  da  by  26.  Ans.  7  mo  3  we  5  ds. 

THE 


44  ARITHMETIC. 

THE  GOLDEN  RULE,  OR  RULE  OF  THREE. 

The  Rule  of  Three  teaches  how  to  find  a  fourth  propor- 
tional to  three  numbers  given  :  for  which  reason  it  is  some- 
times called  the  Rule  of  Proportion.  It  is  called  the  Rule 
of  Three,  because  three  terms  or  numbers  are  given,  to 
find  a  fourth.  And  because  of  its  great  and  extensive  use- 
fulness, it  is  often  called  the  Golden  Rule.  This  Rule  is 
usually  considered  as  of  two  kinds,  namely,  Direct,  and 
Inverse. 

The  Rule  of  Three  Direct  is  that  in  which  more  requires 
more,  or  less  requires  less.  As  in  this  ;  if  3  men  dig  V'  1  yards 
of  trench  in  a  certain  time,  how  much  will  6  men  dig  in  the 
same  time  ?  Here  more  requires  more,  that  is,  6  men,  which 
are  more  than  3  men,  will  also  perform  more  work  in  the 
same  time.  Or  when  it  is  thus  :  if  6  men  dig  42  yards,  how 
much  will  3  men  dig  in  the  same  time  ?  Here  then,  less  re- 
quires less,  or  3  men  will  perform  proportionahly  less  work 
than  6  men,  in  the  same  time.  In  both  these  cases  then,  the 
Rule,  or  the  Proportion,  is   Direct ;  and  the  stating  must  be 

thus,  As  3  :  21  :  :  6  :  42, 
or  thus.  As  6  :  42  :  :  3  :  21, 

But  the  Rule  of  Three  Inverse,  is  when  more  requires  less, 
or  less  requires  more.  As  in  this  :  if  3  men  dig  a  certain 
^  quantity  of  trench  in  14  hours,  in  how  many  hours  will  6 
men  dig  the  like  quantity'?  Here  it  is  evident  that  6  men, 
being  more  than  3,  will  perform  an  equal  quantity  of  work  in 
less  time  or  fewer  hours.  Or  thus  :  if  6  men  perform  a 
certain  quantity  of  work  in  7  hours,  in  how  many  hours  will 
3  men  perform  the  same  ?  Here  less  requires  more,  for  3 
men  will  take  more  hours  than  6  to  perform  the  same  work. 
In  both  these  cases  then  the  Rule,  or  the  Proportion,  is  In- 
verse \  and  the  stating  must  be 

thus.  As  6  ;  14  :  J  3  :    7, 
or  thus.  As  3  :     7  :  :  6  :  14. 

And  in  all  these  statings,  the  fourth  term  is  found,  by 
multiplying  the  2d  and  3d  terms  together,  and  dividing  the  pro- 
duct by  the  1st  term. 

Of  the  three  given  numbers ;  two  of  them  contain  the 
supposition,  and  the  third  a  demand.  And  for  stating  and 
working  questions  of  these  kind  observe  the  following  general 
R«le : 

State 


RULE  OF  THREE.  45 

State  the  question,  by  setting  down  in  a  straight  line  the 
three  given  numbers,  in  the  following  manner,  viz.  so  that 
the  2d  term  be  that  number  of  supposition  which  is  of  the 
same  kind  that  the  an-awer  or  fourth  terra  is  to  be  ;  making  the 
other  number  of  supposition  the  1st  term,  and  the  demanding 
number  the  *5d  term,  when  the  question  is  in  direct  propor- 
tion ;  but  contrariwise,  the  other  number  of  supposition  the 
3d  terra,  and  the  demanding  number  the  1st  term,  when  the 
question  has  inverse  proportion.  * 

Then,  in  both  cases,  multiply  the  2d  and  3d  terms  together, 
and  divide  the  product  by  the  1st,  which  will  give  the  an- 
swer, or  4th  term  sought,  viz.  of  the  same  denomination  as 
the  second  term. 

Note^  If  the  first  and  third  terms  consist  of  different  deno- 
minations, reduce  them  both  to  the  same  :  and  if  the  second 
term  be  a  compound  number,  it  is  mostly  convenient  to  re- 
duce it  to  the  lowest  denomination  mentioned. — If,  after  di- 
vision, there  be  any  remainder,  reduce  it  to  the  next  lower 
denomination,  and  divide  by  the  same  divisor  as  before,  and 
the  quotient  will  be  of  this  last  denomination.  Proceed  in 
the  same  manner  with  all  the  remainders,  till  they  be  re- 
duced to  the  lowest  denomination  which  the  second  admits 
of,  and  the  several  quotients  taken  together  will  be  the  an- 
swer required. 

Note  also,  The  reason  for  the  foregoing  Rules  will  appear, 
when  we  come  to  treat  of  the  nature  of  proportions. — Some- 
times two  or  more  statings  are  necessary,  which  may  always 
be  known  from  the  nature  oi  the  question. 

EXAMPLES. 

I.  UB  yards  of  Cloth  cost  1/  45,  what  will  QQ  yards  cost  ? 

yds   1^       yds    1     s 
As  8  :  1  4  ::  96  :  14  8  the  Answer 
20 

24  ' 

96 


8)  2304 
2,0)   28,8s 

£U  8  Answer. 


Ex.2. 


46  ARITHMETIC. 

Ex.  2.  An  engineer  hailing  raised  lOO  yards  of  a  certain 
work  in  24  days  with  5  men  ;  how  many  men  must  he  em- 
ploy to  finish  a  like  quantity  of  work  in  15  days  ? 

ds   men    ds    men 

As   16  :  6  :  :  24  :  8  Ans. 
5 

•  15)  120  (8  Answer. 
120 

3.  What  will  72  yards  of  cloth  cost,  at  the  irate  of  9  yards 
for  6Z  12s  ?  Ans.  44/  16s, 

4.  A  person's  annual  income  being  146Z ;  how  much  is 
that  per  day  ?  Ans.  Ss, 

5.  If  3  paces  or  common  steps  of  a  certain  person  be  equal 
to  2  yards,  how  many  yards  will  160  of  his  paces  make  ? 

Ans.   106  yds  2  ft. 

6.  What  length  must  be  cut  off  a  board,  that  is  9  inches 
broad,  to  make  a  square  foot,  or  as  much  as  12  inches  in 
length  and  12  in  breadth  contains  ?  Ans.  16  inches. 

7.  If  750  men  re(^uire  22500  rations  of  bread  for  a  month  ; 
how  many  rations  will  a  garrison  of  1200  men  require  ? 

Ans.  36000. 

8.  If  7  cwt  1  qr  of  sugar  cost  26Z  10s  4d ;  what  will  be  the 
price  of  43  cwt  2  qrs  ?  Ans.  159/  25. 

9.  The  clothing  of  a  regiment  of  foot  of  750  men  amount- 
ing to  2831/  5s,-  what  will  the  clothing  of  a  body  of  3500 
inen  amount  to  ?  Ans.  13212/  10s. 

10.  How  many  yards  of  matting,  that  is  3  ft  broad,  will 
cover  a  floor  that  is  27  feet  long  and  20  feet  broad  ? 

Ans.  60  yards. 

11.  What  is  the  value  of  6  bushels  of  coals,  at  the  rate  of 
1/  Ms  6f/  the  chaldron  ?  Ans.  5s  9d, 

12.  If  6352  stones  of  3  feet  long  complete  a  certain  quan- 
tity of  walling  ;  how  many  stones  of  2  feet  long  will  raise  a 
like  quantity  ?  Ans.  9528. 

13.  What  must  be  given  for  a  piece  of  silver  weighing 
73  lb  5  oz  15  dwts,  at  the  rate  of  5s  9d  per  ounce  ? 

Ans.  253/  10s  Of  (^. 

14.  A  garrison  of  536  men  having  provision  for  12  months  ; 
how  long  will  those  provisions  last,  if  the  garrison  be  increase* 
to  1124  men  ?  Ans.  174  days  and  xfl?- 

15.  What  will  be  the  tax  upon  763/  15s,  at  the  rate  of 
3s  Qd  per  pound  sterling  ?  Ans.  133/  13s  l-^rf. 

16.  A 


RULE  OF  THREE.  47 

16.  A  certain  work  being  raised  in  12  days,  by  working  4 
hours  each  day  ;  how  long  would  it  have  been  in  raising  by 
working  6  hours  per  day  ?  Ans.  8  days. 

17.  What  quantity  of  corn  can  I  buy  for  90  guineas,  at  the 
rate  of  6s  the  bushel  ?  Ans  39  qrs  3  bu. 

18.  A  person,  failing  in  trade,  owes  in  all  977/ ;  at  which 
time  he  has,  in  money,  goods,  and  recorerable  debts,  .420Z  6s 
^id  ;  now  supposing  these  things  delivered  to  his  creditors, 
how  much  will  they  get  per  pound  ?  Ans.  85  7Arf. 

19.  A  plain  of  a  certain  extent  having  supplied  a  body  of 
3000  horse  with  forage  for  18  days  ;  then  how  many  days 
Would  the  same  plain  have  supplied  a  body  of  2000  horse  ? 

Ans  27  days. 

20.  Suppose  a  gentleman's  income  is  600  guineas  a  year, 
and  that  he  spends  25s  6d  per  day,  one  day  with  another  ; 
how  much  will  he  have  saved  at  the  year's  end  ? 

Ans.  164/  125  6d. 

21.  What  cost  30  pieces  of  lead,  each  weighing  1  cwt  121b, 
at  the  rate  of  16s  4d  the  cwt  ?  Ans.  27/  2s  6c/. 

22.  The  governor  of  a  besieged  place  having  provision  for 
64  days,  at  the  rate  of  l^lb  of  bread  ;  but  being  desirous  to 
prolong  the  siege  to  80  days,  in  expectation  of  succour,  in 
that  case  what  must  the  ration  of  bread  be  ?  Ans,  1 J^  lb. 

23.  At  half  a  guinea  per  week,  how  long  can  1  be  boarded 
for  20  pounds  ?  Ans.  38  J|^  wks. 

24.  How  much  will  75  chaldrons  7  bushels  of  coals  come 
to,  at  the  rate  of  1/  I3s  6d  per  chaldron  ? 

Ans.  126/  19s  Oi</. 

25.  If  the  penny  loaf  weigh  8  ounces  when  the  bushel  of 
'  wheat  costs  7s  3c/,  what  ought  the  penny  loaf  to  weigh  when 

the  wheat  is  at  8s  4d  ?  Ans.  6  oz  15y3_e_  dr. 

26.  How  much  a  year  will  173  acres  2  roods  14  poles  of 
land  givci  at  the  rate  of  1/  7s  8c/  per  acre  ? 

Ans.  240/  2s  l^^d. 

27.  To  how  much  amounts  73  pieces  of  lead,  each  weigh- 
ing 1  cwt  3  qrs  7  lb,  at  10/  4s  per  fother  of  I9i  cwt  ? 

Ans.  69/  4s  2d  l^f  q, 

28.  How  many  yards  of  stuff,  of  3  qrs  wide,  will  line  a 
cloak  that  is  If  yards  in  length  and  2>\  yards  wide  ? 

Ans.  8  yds  0  qrs  2|  nL 

29.  If  5  yards  of  cloth  cost  145  2c/,  what  must  be  given  for 
9  pieces,  containing  each  21  yards  1  quarter  ? 

Ans.  27/  Is  lOic/, 

30.  If  a  gentleman's  estate  be  worth  2107/  12s  a  year ; 
what  may  he  spend  per  day,  to  save  500/  in  the  year  ? 

Ans.  4/.8»  I^Vjc/. 
31.  Wanting 


48  ARITHMETIC. 

31.  Wanting  just  an  acre  of  land  cut  off  from  a  piece 
which  is  13i  poles  in  breadth,  what  length  must  the  piece  be  ? 

Ans.  11  po  ^  yds  2  ft  Oif  in. 

32.  At  7s  9^d  per  yard,  what  is  the  value  of  a  piece  of 
cloth  containing  63  ells  English  1  qu.  Ans.  251  18s  Ifrf. 

33.  If  the  carriage  of  5  cwt  141b  for  96  miles  be  1/  12s  6d; 
how  far  may  i  have  3  cwt   1  qr  carried  for  the  same  money  ? 

Ans.  151  m  3  fur  3J^  pol. 

34.  Bought  a  silver  tankard,  weighing  1  lb  7  oz  14  dwts  ; 
what  did  it  cost  me  at  6s  4d  the  ounce  ?  Ans.  61  4$  9^d, 

36.  What  is  the  half  year's  rent  of  547  acres  of  land,  at 
15s  6d  the  acre  ?  Ans.  21 U  19s  3d. 

36.  A  wall  that  is  to  be  built  to  the  height  of  36  feet,  was 
raised  9  feet  high  by  16  men  in  6  days  ;  then  how  many  men 
must  be  employed  to  finish  the  wall  in  4  days,  at  the  same 
rate  of  working  ?  Ans.  72  men 

37.  What  will  be  the  charge  of  keeping  20  horses  for  a 
year,  at  the  rate  of  H^rf  per  day  for  each  horse  ? 

Ans.  441/  Os  lOd, 

38.  If  18  ells  of  stuff  that  is  f  yard  wide,  cost  39s  6d  ; 
what  will  50  ells,  of  the  same  goodness,  cost,  being  yard  wide? 

Ans.  71  6s  3f  |rf.. 

39.  How  many  yards  of  paper  that  is  30  inches  wide,  will 
hang  a  room  that  is  20  yards  in  circuit  and  9  feet  high. 

Ans.  72  yards. 

40.  If  a  gentleman's  estate  be  worth  3«4Z  16s  a  year,  and 
the  land-tax  be  assessed  at  2s  d^d  per  pound',  what  is  his  net 
annual  income  ?  Ans.  3311  Is  9\d, 

41.  The  circumference  of  the  earth  is  about  25000  miles  ; 
at  what  rate  per  hour  is  a  person  at  the  middle  of  its  surface 
carried  round,  one  whole  rotation  being  made  in  23  hours 
66  minutes  ?  Ans.  1044yV3V  miles. 

42.  If  a  person  drink  20  bottles  of  wine  per  month,  when 
it  costs  8s  a  gallon ;  how  many  bottles  per  month  may  he 
drink,  without  increasing  the  expence,  when  wine  costs  10s, 
the  gallon?  Ans.  16  bottles. 

43.  What  cost  43  qrs  6  bushels  of  corn,  at  IZ  8s  &d  the 
quarter.  Ans.  62Z  3s  Sfd. 

44.  How  many  yards  of  canvas  that  is  ell  wide,  will  line 
60  yards  of  say  that  is  3  quarters  wide  ?  Ans.  30  yds. 

45.  If  an  ounce  of  gold  cost  4  guineas,  what  is,the  value 
of  a  grain  ?  Ans.  '^j'^d, 

46.  If  3  cwt  of  tea  cost  401  12s;  at  how  much  a  pound 
must  it  be  retailed,  to  gain  10/  by  the  whole  ?        Ans.  S^^^s,- 


COMPOUND 


f  49] 


COMPOUND  PROPORTION. 


©OMPOUND  Proportion  shows  how  to  resolve  such  quesj- 
tions  as  require  two  or  more  statings  by  Simple  Proportion ; 
and  these  may  be  either  Direct  or  Inverse. 

In  these  questions,  there  is  always  given  an  odd  number  of 
terras,  either  five,  or  seven,  or  nine,  kc.  These  are  distinr- 
guished  into  terms  of  supposition,  and  terms  of  demand, 
there  being  always  one  term  more  of  the  former  than  of  the 
latter,  which  is  of  the  same  kind  with  the  answer  sought; 
The  method  is  thus  : 

Set  down  in  the  middle  place  that  term  of  supposition 
which  is  of  the  same  kind  with  the  answer  sought. — Take 
one  of  the  other  terms  of  supposition,  and  one  of  the  demand- 
ing terms  which  is  of  the  same  kind  with  it ;  then  place  one 
of  them  for  a  first  term,  and  the  other  for  a  third,  according 
to  the  directions  given  in  the  Rule  of  Three. — Do  the  same 
with  another  term  of  supposition,  and  its  corresponding  de- 
manding term  ;  and  so  on  if  there  be  more  terms  of  each 
kind  ;  setting  the  numbers  under  each  other  which  fall  all  on 
the  left-hand  side  of  the  middle  term,  and  the  same  for  the 
others  on  the  right-hand  side. — Then,  to  work 

By  several  Operations. — Take  the  two  upper  terms  and 
the  middle  term,  in  the  same  order  as  they  stand,  for  the  first 
Rule-of- Three  question  to  be  worked,  whence  will  be  found 
a  fourth  term.  Then  take  this  fourth  number,  so  found,  for 
the  middle  term  of  a  second  Rule-of-Three  question,  and  the 
next  two  under  terms  in  the  general  stating  in  the  same 
order  as  they  stand,  finding  a  fourth  term  for  them.  And  so 
on,  as  far  as  there  are  any  numbers  in  the  general  stating, 
making  always  the  fourth  number,  resulting  from  each  simple 
stating,  to  be  the  second  term  in  the  next  following  one. 
So  shall  the  last  resulting  number  be  the  answer  to  the 
question. 

By  one  Operation. — Multiply  together  all  the  terms  stand- 
ing under  each  other,  on  the  left-hand  side  of  the  middle 
term  ;  and  in  like  manner,  multiply  together  all  those  on  the 
right-hand  side  of  it.  Then  multiply  the  middle  term  by 
the  latter  product,  and  divide  the  result*  by  the  former  pro- 
duct ;  so  shall  the  quotient  be  the  answer  sought. 

V©L.  1.  8  ^.     EXAMPIiES 


50 


ARITHMETIC. 
EXAMPLES. 


1.  How  many  men  can  complete  a  trench  of  135  yards 
long  in  8  days,  when  16  men  can  dig  54  yards  in  6  days  ? 


General  Stating. 

yds.    54 
days      8 

16 

:  :  135  yds 
6  days 

432 

810- 
16 

32) 

4860 

81      men 

4 

12960  (30  Ans.  by  one  operation 
1296 

The  same  by  two  Operations. 


54 

1st. 
16  :  :  135 
16 

40 

2d. 
As  8  :  40  :  :  6  :  30 
6       , 

810 
135 

8)  240  (30  Ans. 
24 

54)2160 
216 

(40^ 

0 

2.  If  100/  in  one  year  gain  bl  interest,  what  will  be  tb« 
interest  of  750/  for  7  years  ?  Ans.  262/ 10*. 

3.  If  a  family  of  8  persons  expend  200/  in  9  months  ;  how 
much  will  serve  a  family  of  18  people  12  mo&tfas  ? 

Ans.  300/^ 

4.  If  27s  be  the  wages  of  4  men  for  7  days  ;  what  will  be 
the  wages  of  14  men  for  10  days  ?  Ans.  6/  16j. 

6.  If  a  footman  travel  130  miles  in  3  days,  when  the  days 
are  12  hours  long  ;  in  how  many  days,  of  10  hours  each, 
may  he  travel  360  miles  ?  /  Ans.  9|f  days. 

Ex.  6. 


VULGAR  FRACTIONS.  61 

Ex.  6.  If  120  bushels  of  corn  can  serve  14  horses  56  days  ; 
how  many  days  will  94  bushels  serve  6  horses  ? 

Ans.  102^^  days. 

7.  If  3000  lb.  of  beef  serve  340  men  16  days  ;  how  many 
lbs  will  serve  120  men  for  26  days  ?        Ans.  1764  lb  1 1  if  oz. 

8.  If  a  barrel  of  beer  be  sufficient  to  last  a  family  of  8  per* 
sons  12  days  ;  how  many  barrels  will  be  drank  by  16  persons 
in  the  space  of  a  year  ?  Ans.  60f  barrels. 

9.  If  180  men  in  6  days,  of  10  hours  each,  can  dig  a  trench 
200  yards  long, -3  wide,  and  2  deep  ;  in  how  many  days,  of  8 
hours  longj  will  100  men  dig  a  trench  of  360  yards  long,  4  wide, 
and  3  deep  ?  Ans.  15  days. 


OF  VULGAR  FRACTIONS. 


A  Fraction,  or  broken  number,  is  an  expression  of  a  part, 
or  some  parts,  of  something  considered  as  a  whole. 

It  is  denoted  by  two  numbers,  placed  one  below  the  other, 
with  a  line  between  them  : 

3  numerator      ^  . 
Thus,  -          '              >  which  is  named  3-fourths. 

4  denominator  ) 

The  denominator,  or  number  placed  below  the  line,  shows 
how  many  equal  parts  the  whole  quantity  is  divided  into  ; 
and  it  represents  the  Divisor  in  Division. — And  the  Numera- 
tor, or  number  set  above  the  Hne,  shows  how  many  of  these 
parts  are  expressed  by  the  Fraction  ;  being  the  remainder 
after  division. — Also,  both  these  numbers,  are  in  general, 
named  the  Terms  of  the  Fraction. 

Fractions  are  either  Proper,  Improper,  Simple,  Compound, 
or  Mixed. 

A  proper  Fraction,  is  when  the  numerator  is  less  than  the 
denominator  ;  as,^,  or  |,  or  f ,  &c. 

An  Improper  Fraction,  is  when  the  numerator  is  equal  to, 
or  exceeds,  the  denominator  ;  as,  f,  or  f,  or|,  &c. 

A  Simple  Fraction,  is  a  single  expression,  denoting  any  num- 
ber of  parts  of  the  integer  ;  as,  |,  or  #. 

A  Compound  Fraction,  is  the  fraction  of  a  fraction,  or  seve- 
ral fractions  connected  with  the  word  of  between  them  :  as, 
i  of  |,orf  off  of  3,  &c. 

A  Mixed  Number,  is  composed  of  a  whole  number  and  a 
fraction  together  ;  as,  3|,  or  12f ,  &c. 

A  whole 


i^2-  ARITHMETIC, 

A  whole  or  integer  niioober  may  be  expressed  like  a  frac- 
tion, by  writing  I  below  it,  as  a  denominator  ;  so  3  is  f ,  or  4 
is  f ,  &c. 

A  fraction  denotes  division  ;  and  its  value  is  equal  to  the 
quotient  obtained  by  dividing  the  numerator  by  the  denomina- 
tor ;  so  *^  is  equal  to  3,  and  ^-o  is  equal  to  4. 

Hence  then,  if  the  numerator  be  less  than  the  denominator, 
the  value  of  the  fraction  is  less  than  1.  But  if  the  numerator 
be  the  same  as  the  denominator,  the  fraction  is  just  equal  to  1. 
And  if  the  numerator  be  greater  than  the  denominator,  the  frac- 
^on  is  greater  than  1 . 


REDUCTION  OF  VULGAR  FRACTIONS. 

Reduction  of  Vulgar  Fractions,  is  the  bringing  them  out  of 
one  form  or  denomination  into  another  ;  commonly  to  prepare 
them  for  the  operations  of  Addition,  Subtraction,  &c.  of  which 
there  are  several  cases. 

\ 
PROBLEM. 

'To  find  the  Greatest  Common  Measure  of  Two  or  mqre  Num- 
bers. 

The  Common  Measure  of  two  or  more  numbers,  is  that 
number  which  will  divide  them  both  without  remainder  ;  so, 
3  is  a  common  measure  of  18  and  24  ;  the  quotient  of  the 
former  being  6,  and  of  the  latter  8.  And  the  greatest  num- 
ber that  will  do  this,  is  the  greatest  common  measure  :  so  6 
is  the  greatest  common  measure  of  18  and  24  ;  the  quotient  of 
of  the  former  being  3,  and  of  the  latter  4,  which  will  not  both 
divide  farther. 


nULE: 

If  there  be  two  numbers  only  ;  divide  the  greater  by  the 
less  ;  then  divide  the  divisor  by  the  remainder  ;  and  so  on, 
dividing  always  the  last  divisor  by  the  last  remainder,  till  no- 
thing remains  ;  so  shall  the  last  divisor  of  all  be  the  greatest 
common  measure  sought. 

When  there  are  more  than  two  numbers,  find  the  greatest 
common  measure  of  two  of  them,  as  before  ;  then  do  the 
same  for  that  common  measure  and  another  of  the  numbers  ; 

and 


REDUCTION  OP  VULGAR  FRACTIONS.  63 

and  so  on,  through  all  the  numbers  ;  so  will  the  greatest  com- 
mon measure  last  found  be  the  answer. 

If  it  happen  that  the  common  measure  thus  found  is  1  ; 
then  the  numbers  are  said  to  be  incommensurable,  or  not 
having  any  common  measure. 


EXAMPLES. 

1.  To  find  the  greatest  common  measure  of  1908,  936, 
and  630. 

936)  1908  (2  So  that  36  is  the  greatest  common 

1872  measure  of  1908  and  936. 

36)  936  (26  Hence  36)  630  (17 

72  36 

216  270 

216  252 


18)36  (2 


Hence  then  18  is  the  answer  require^. 

2.  What  is  the  greatest  common  measure  of  246  and  372  1 

Ans.  6. 

3.  What  is  the  greatest  common  measure  of  324,  612,  and 
1'032  ?  Ans.  12. 


CASE  L 

To  Abbreviate  or  Reduce  Fractions  to  their  Lowest  Terms. 

*  Divide  the  terms  of  the  given  fraction  by  any  number 
that  will  divide  them  without  a  remainder  ;  then  divide  these 

quotients 


*  That  dividing  both  the  terms  of  ihe  fraction  by  the  same  number, 
whatever  it  be,  will  give  another  fraction  equal  to  Ihe  former,  is  evi- 
dent. And  when  these  divisions  are  performed  as  often  as  can  be 
done,  or  when  the  common  divisor  is  the  greatest  possible,  the  terms 
of  the  resulting  fraction  must  be  the  least  possible. 

Note  1.  Any  number  ending  with  an  even  number,  or  a  cipher,  is 
divisible,  or  can  be  divided,  by  2. 

2.  Any  number  ending  with  5,  or  0,  is  divisible  by  5. 

3.  If 


64  ARITHMETIC. 

quotients  again  in  the  same  manner  ;  and  so  on,  till  it  appears 
that  there  is  no  number  greater  than  1  which  will  divide 
them  ;  then  the  fraction  will  be  in  its  lowest  terms. 

Or,  divide  both  the  terms  of  the  Fraction  by  their  greatest 
common  measure  at  once,  and  the  quotients  will  be  the  terms 
of  the  fraction  required,  of  the  same  value  as  at  first. 


EXAMPLES. 

.  Reduce  flf  to  its  least  terms. 
lif  =  H  =  f  I  =  If  =  I  =  f  •  the  answer.  . 
Or  thus  : 
16)  288(1  Therefore  72  is  the  greatest  common 

216  measure;  and  72)  fif  =f  the  An- 

swer,  the  same  as  before. 

72)  216  (3 
216 

2.  Reduce 


3.  If  the  right-hand  place  of  any  number  be  O.tbe  whole  is  divisible 
by  10  ;  if  there  be  two  ciphers,  it  is  divisible  by  100  ;  if  three  ciphers 
by  1000 :  and  so  on  ;   which  is  only  cutting  off  those  ciphers. 

4.  If  the  two  right-hand  figures  of  any  number  be  divisible  by  4,  tlie 
whole  is  divisible  by  4.  And  if  the  three  right-hand  figures  be  divisi- 
ble by  8,  the  whole  is  divisible  by  8.     And  so  on. 

5.  If  the  sum  of  the  digits  in  any  number  be  divisible  by  S,  or  by  9, 
the  whole  is  divisible  by  3»  or  by  9. 

6.  If  the  right  hand  digit  be  even,  and  the  sum  of  all  tha  digits  be 
divisible  by  6,  then  the  whole  is  divisible  by  6. 

7.  A  number  is  divisible  by  11,  when  the  sum  of  the  1st,  3d,  5th, 
Sec  or  all  the  odd  places,  is  equal  to  the  sum  of  the  2d,  4th,  6th,  &c.  or 
of  all  the  even  places  of  digits. 

8.  If  a  number  cannot  be  divided  by  some  quantity  less  than,  the 
square  root  of  the  same,  that  number  is  aprimey  or  cannot  be  dindcd 
by  any  number  whatever. 

9.  All  prime  numbers,  except  2  and  5,  have  either  1,  3,  7,  or  9, 
in  the  place  of  units  ,  and  all  other  numbers  are  composite,  or  can  be 
divideo. 

iO.  When 


REDUCTION  OF  VULGAR  FRACTIONS.  65 

2.  Reduce  Iff  to  its  lowest  terms.  Ans.  a. 

3.  Reduce  ^f  f  to  its  lowest  terms.  Ans.  f . 

4.  Reduce  |f  f  to  its  lowest  terms.  Ans.  |. 


CASE  K. 

To  Reduce  a  Mixed  Number  to  its  Equivalent  improper  Fraction, 

*  Multiply  the  integer  or  whole  number  by  the  deno- 
minator of  the  fraction,  and  to  the  product  add  the  numera- 
tor ;  then  set  that  sum  above  the  denominator  for  the  frac- 
tion required. 

EXAMPLES. 

1.  Reduce  23|  to  a  fraction. 
23 
5 

115  Ory 

2  (23X6)4-2  117 

' = ,  the  Answer. 

117  5  5 


2.  Reduce  12^  to  a  fraction.  Ans.  ^s, 

3.  Reduce  14yV  to  a  fraction.  Ana.  y/. 

4.  Reduce  183^5_  to  a  fraction.  Ans.  ^i**. 


10.  When  narabers,  with  the  sign  of  addition  or  subtraction  be- 
tween them,  are  to  be  divided  by  any  number,  then  each  of  those 

10+8-4 
numbers  must  be  divided  by  it    Thus  =54-4  —2=7. 

2 
H.  But  if  the  numbers  have  the  sign  of  multiplication  between 
them,  only  one  of  them  must  be  divided.     Thus, 
10X8X3        10X4X3       10X4x1        10X2x1        20 

6X2  6X1       .      2X1  1X1  1 

*  This  is  no  more  than  first  muhrplying  a  quantity  lyy  some  num- 
ber, and  then  dividing  the  result  back  again  by  the  same  i  which  it  is 
evident  does  not  alter  the  value  j  for  any  fraction  represents  a  division 
of  the  numerator  by  the  denominator. 

GASE 


5<j  ARITHMETIC 


CASE  m. 

To  Reduce  an  Improper  Fraction  to  it    Equivalent  Whole  oj- 
Mixed  Number. 

*  Divide   the    numerator    hy  the  denominator,    and   the 
quotient  will  be  the  whole  or  mixed  number  sought. 

EXAMPLES. 

1 .  Reduce  y  to  its  equivalent  number. 

Here  y  or  12  4-  3  =  4,  the  Answer. 

2.  Reduce  y  to  its  equivalent  number. 

Here  y  or  15  -f-  7  =  24,  the  Answer. 

3.  Reduce  \%^  to  its  equivalent  number* 
Thus   17)  749  (44yV 

68 


69  So  that  VV  ^  44xV,  the  Answer. 

68 


4.  Reduce  Y  to  its  equivalent  number.  Ans.  8. 

5.  Reduce  ^ff^  to  its  equivalent  number.  Ans.  54i^f . 

6.  Reduce  3^1.8  tQ  its  equiyaient  number.         Ang.  171|4. 

*  CASE  IV. 

To  Reduce  a  Whole  Number  to  an  Equivalent  Fraction,  having  a 
Given  denominator, 

t  Multiply  the  whole  number  by  the  given  denominator  : 
then  set  the  product  over  the  said  denominator,  and  it  will 
form  the  fraction  required. 


*  This  rule  is  evidently  the  reverse  of  the  former  ;  and  the  reason 
of  it  is  manifest  from  the  nature  of  Common  Division, 

f  Multiplication  and   Division  being  here  equally  used,  the  result 
must  b&  the  same  as  the  quantity  first  proposed. 

EXAMPLES, 


REDUCTION  OF  VULVAR  FRACTIONS,    67 

EXAMPLES. 

1.  Reduce  9  to  a  fractioa  whose  denominator  shall  be  7. 

Here  9  X  7  =  63  :  then  V  ^^  the  Answer  ; 
For  Y  =  63  -r-  7  =  9,  the  Proof. 

2.  Reduce  12  to  a  fraction  whose  denominator  shall  be  13. 

Ans.  V¥- 

3.  Reduce  27  to  a  fraction  whose  denominator  shall  be  IK 

Ans.  Vt^- 

CASE  V. 

To  Reduce  a  Compound  Fraction  to  an  Equivalent  Simple  One. 

*  Multiply  all  the  numerators  together  for  a  numerator, 
and  all  the  denominators  together  for  a  denominator,  and 
they  will  form  the  simple  fraction  sought. 

When  part  of  the  compound  fraction  is  a  whole  or  mixed 
number,  it  must  first  be  reduced  to  a  fraction  by  one  of  the 
former  cases. 

And,  when  it  can  be  done,  any  two  terms  of  the  fraction 
may  be  divided  by  the  same  number,  and  the  quotients  used 
instead  of  them.  Or,  when  there  are  terms  that  are  common, 
they  may  be  ofliitted,  or  cancelled. 

EXAMPLES. 

1.  Reduce  x  of  f  of  |  to  a  simple  fraction. 

1X2X3       6         1 
Here =  —  =  — ,  the  Answer. 

2  X  3  X  4      24        4 

ix^xgr     1 

Or,   =  — ,  by  cancelling  the  2*s  and  3's. 

^X^X4       4 

*  The  truth  of  this  Rule  may  be  shown  as  follows :  Let  the  com- 
pound fraction  be  |  of  1.  Now  X  of  ^  is  1  -f-  3,  which  is  _5_  consequent- 
ly 2  of  s  will  be  -5-  X  2  or  4.^  ;  that  is  the  numerators  are  multiplied 
together,  and  also  the  denominatorsi  as  in  the  Rule.  When  the  com- 
pound fraction  consists  of  more  than  two  single  ones  ;  having  first  re- 
duced two  of  them  as  above,  then  the  resulting  fraction  and  a  third 
will  be  the  same  as  a  compound  fraction  of  two  parts  j  and  so  on  to 
the  last  of  all. 

2.  Reduce 
Vol,  L  S 


58  ARITHMETIC. 

2.  Reduce  |  of  |  of  \{  to  a  simple  fraction. 

2  X  3  X  10        60       12        4. 

Here — =  —  =  —  =  — ,  the  Answer. 

3  X  6  X  11       165     33        11 

2  X^Xi:jef       '4  " 

Or, =  — ,  the  same  as  before,  by  cancelling 

^X^X  11         11 
the  3's,  and  dividing  by  6's. 

3.  Reduce  ^  of  a  to  a  simple  fraction.  Ans.  if. 

4.  Reduce  |  of  |  of  f  to  a  simple  fraction.  Ans.  f. 

5.  Reduce  f  of  |  of  3i  to  a  simple  fraction.  Ans.  |. 

6.  Reduce  f  of  |  of  |  of  4  to  a  simple  fraction.  Ans.  |. 

7.  Reduce  2  and  |  of  |  to  a  fraction.  Ans.  |. 

CASE  VI. 

To  Reduce  Fractions  of  Different  Denominators ^  to  Equivalent 
Fractions  having  a  Common  Denominator, 
*  Multiply  each  numerator  by  all  the  deriominators  ex- 
cept its  own,  for   the  new  numerators  :   and  multiply  all  the 
denominators  together  for  a  common  denominator. 

Note,  It  is  evident  that  in  this  and  several  other  operations, 
when  any  of  the  proposed  quantities  are  integers,  or  mixed 
umbers,  or  compound  fractions,  they  must  first  be  reduced, 
ny  their  proper  Rules,  to  the  form  of  simple  fractions, 
b 

EXAMPLES. 

1.  Reduce  i,  |,  and  f ,  to  a  common  denominator. 

1  X3X4  =  12  the  new  numerator  for  i 

2  X  2  X  4  =  16  ditto  | 

3  X  2  X  3  =  18  ditto  | 
2  X  3  X  4  =  24  the  common  denominator. 

Therefore  the  equivalent  fractions  are  ^|,  |f ,  and  if. 
Or  the  whole  operation  of  multiplying  may  be  best  per- 
formed mentally,  only  setting  down  the   results  and    given 

frarfinns   thn«5  •    JL     3.     3      =     it     16     J..8      =:::    __6_     _8_     _9_   V^v 

JIdCUUUS     IIIUS   .     2*     3^»     4J      2ii     24»     24'     I2>     T2>     T2     "J 

abbreviation. 

2.  Reduce  f  and  |  to  fractions  of  a  common  denominator. 

Ans.  -1^.  -3^^ 

■ 1 • • ■ — ■ ■^ 

*  This  is  evidently  no  more  than  multiplying  each  numerator  an'l  its 
denominator  by  the  same  quantity,  and  consequently  the  value  of  the 
fraction  is  not  altered. 

See  the  note  respecting  the  least  common  denominator  at  the  end 
of  vol.  T. 

3.  Reduce 


REDUCTION  OF  VULGAR  FRACTIONS.  6& 

3.  Reduce  ^,4,  and  |,  to  a  common  denominator. 

Ans.  a.  ih  U' 

4.  Reduce  4,  24,  and  4  to  a  common  denominator. 

Ans.H^if,  W- 
J^ote  I.    When    the  denoririnators  of  two  given  fractions 
have  a  common  measure,  let  them   be  divided   by  it  ;  then 
multiply  the  terms  of  each  given  fraction  by  the  quotient  aris- 
ing, from  the  other's  denominator. 

Ex,  /^,  and  /^  =  jtV  and  yVs*  ^y  multiplying  the  former 
5  7  ,  by  7,  and  the  latter  by  6. 

2.  When  the  less  denominator  of  two  fractions  exactly  di- 
vides the  greater,  multiplyithe  terms  of  that  which  has  the  less 
denominator  by  the  quotient. 

Ex.  f  and  y\  =  j\  and  y\,  by  mult,  the  former  by  2. 

2 

3.  When  more  than  two  fractions  are  proposed,  it  is  some- 
times convenient,  first  to  reduce  two  of  them  to  a  common  deno- 
minator ;  then  these  and  a  third  ;  and  so  on  till  they  be  all  re- 
duced to  their  least  common  denominator. 

Ex,  f  and  f  and  I  =  I  and  f  and  |  =  if  and  ||  and  f  f 


CASE  vir. 


To  Jind  the   value  of  a  Fraction  in  Paris  of  the  Integer. 

MuLTiPBY  the  integer  by  the  numerator,  and  divide  the  pro- 
duct by  the  denominator,  by  Compound  Multiplication  and  Di- . 
vision,  if  the  integer  be  a  compound  quantity. 

Or,  if  it  be  a  single  integer,  multiply  the  numerator  by  the 
parts  in  the  next  inferior  denomination,  and  divide  the  pro- 
duct by  the  denominator.  Then,  if  any  thing  remains,  mul- 
tiply it  by  the  parts  in  the  next  inferior  denomination,  and  di- 
vide by  the  denominator  as  before  ;  and  so  on  as  far  as  neces- 
sary ;  so  shall  the  quotients,  placed  in  order,  be  the  value  of 
the  fraction  required*. 


♦  The  numerator  of  a  fraction  being  considered  as  a  remainder,  ia 
Division,  and  the  denominator  as  the  divisor,  this  rule  is  of  the  same 
nature  as  Compound  Division,  or  the  valuation  of  remainders  in  the 
Rule  «f  Three*  before  explained. 

EXAMPLES. 


6i» 


ARITHMETIC: 


EXAMPLES. 


1.  What  is  the  f  of  2/ 6s  ? 
By  the  former  part  of  the  Rule 
21  6s 
4 


Ans. 


6)  9  4 
1/  16s9d2|y. 


2.  What  is  the  value  of  |  of  U  ? 
By  the  2d  part  of  the  Ruje, 
2 
20 

3)  40  (13s  4d  Ans. 

1 

12 

3)  n{Ad 


3.  Find  the  value  of  f  of  a  pound  sterling.         Ans.  7s  6d. 

4.  What  is  the  value  of  |  of  a  guinea  ?  Ans.  4s  8d: 

5.  What  is  the  value  of  f  of  a  half  crown  ?   Ans.  Is  10i(|. 

6.  What  is  the  value  of  f  of  4s  lOd  1  Ans.  Is  1  l^d 

7.  What  is  the  value  of  f  lb  troy  ?  Ans.  9  oz  12  dwta. 

8.  What  is  the  value  of  j\  of  a  cwt  ?  Ans.  1  qr  7  lb. 

9.  What  is  the  value  of  f  of  an  acre  ?     Ans.  3  ro.  20  po. 
10.  What  is  the  value  of  f_  pf  a  day  ?     Ans.  "^  hrs  12  min. 

CASE  VIII. 

To  Reduce  a  Fraction  from  one  Denomination  to  another. 

*  Consider  how  many  of  the  less  denomination  make  one 
of  the  greater  ;  then  multiply  the  numerator  by  that  number, 
if  the  reduction  be  to  a  less  name,  but  multiply  the  denomina- 
tor, if  to  a  greater. 

EXAMPLES. 

1.  Reduce  f  of  a  pound  to  the  fraction  of  a  penny, 

I  X  V  X  V^  =  4|»  =  H%  the  Answer. 


*  rhis  is  the  same  as  the  Rule  of  Reduction  in  whole  numbevs  from 
one  denomination  to  another. 


S.  Reduce 


ADDITION  OF  VULGAR  FRACTIONS.  61 

2.  Reduce  ^^  of  a  penny  to  the  fraction  of  a  pound. 

4  X  y'^  X  2^  =  ok  the  Answer. 

3.  Reduce  ^^l  to  the  fraction  of  a  penny.  Ans.  ^^d, 

4.  Reduce  f q  to  the  fraction  of  a  pound.  Ans.  -^is-o* 

5.  Reduce  f  cwt  to  the  fraction  of  a  lb.  Ans.  ^^  . 

6.  Reduce  f  dwt  to  the  fraction  of  a  lb  troy.  Ans  ji^. 

7.  Reduce  f  crown  to  the  fraction  of  a  guinea.       Ans.  /g^. 

8.  Reduce  |  half-crown  to  the  fract.  of  a  shilling.    Ans.  f  f . 

9.  Reduce  2s  Qd  to  the  fraction  of  a  £.  Ans.  |. 
10.  Reduce  17s  Id  ^q  to  the  fraction  of  a  £. 


ADDITION  OF  VULGAR  FRACTIONS. 

If  the  fractions  have  a  common  denominator ;  add  all  the 
numerators  together,  then  place  the  sum  over  the  common 
denominator,  and  that  will  be  the  sum  of  the  fractions 
required. 

*  If  the  proposed  fractions  have  not  a  common  denomina- 
tor, they  must  be  reduced  to  one.  Also  compound  fractions 
must  be  reduced  to  simple  ones,  and  fractions  of  different 
denominations  to  those  of  the  same  denomination.  Then 
add  the  numerators  as  before.  As  to  mixed  numbers,  they 
may  either  be  reduced  to  Improper  fractions,  and  so  added 
with  the  others  ;  or  else  the  fractional  parts  only  added,  and 
the  integers  united  afterwards. 


*  Before  fractions  are  reduced  to  a  common  denominator,  they  are 
quite  dissimilar,  as  much  as  shillings  and  pence  are,  and  therefore 
cannot  be  incorporated  with  one  another,  any  more  than  these  can. 
But  when  they  are  reduced  to  a  common  denominator,  and  made  parts 
of  the  same  tiling,  their  sum,  or  difference,  may  then  be  as  properly 
expressed  by  the  sum  or  difference  of  the  numerators,  as  the  sum  or 
difference  of  any  two  quantities  whatever,  by  the  sum  or  difference  of 
their  individuaJs.  Whence  the  reason  of  the  Rule  is  manifest,  both 
tor  Addition  and  Subtraction. 

When  several  fractions  are  to  be  collected,  it  is  commonly  best  first 
to  add  two  of  them  together  that  most  easily  reduce  to  a  common  de- 
nominAtor  ;  then  add  their  sum  and  third,  and  so  on. 

EXAMPLES. 


g2  ARITHMETIC. 

EXAMPLES. 

1.  To  add  f  and  f  together. 

Here  |  +  j  =  j  =  If »  the  Answer. 

2.  To  add  |  and  |  together. 

3  4-  i  =  ±8  J_  2  5   =  4  3  :=;  1  1.3    the  Answpr 

3.  To  add  f  and  7i  and  |  of  |  together. 

I  +  7i  +  iof  f  =1  -f  y  +i  =1 4-  V  +f  =  ¥=8f 

4.  To  add  ^  and  f  together.  Ans.  If. 
6.  To  add  |  and  f  together.                                    Ans.  m. 

6.  Add  f  and  j\  together.  Ans.  i^. 

7.  What  is  the  sum  of  |  and  f  and  f  ?  Ans.  1|^|. 

8.  What  is  the  sum  of  |  and  f  and  2J-  ?  Ans.  Sf-f. 

9.  What  is  the  sum  of  |  and  |^  of  i  and  9/^  ?  Ans.  lOgV- 

10.  What  is  the  sum  of  |  of  a  pound  and  f  of  a  shilling  ? , 

Ans.  »|5s  or  13s  lOd  2^q. 

11.  What  is  the  sum  of  |  of  a  shilling  and  y*j  of  a  penny  9 

Ans.  y-^d  or  Id  lj\q, 
.12.  What  is  the  sum  of  i  of  a  pound,  and  |  of  a  shilling, 
and  j%  of  penny  ?  Ans.  Hof^  or  3*  ^^  Hf  ?• 


SUBTRACTION  OF  VULGAR  FRACTIONS. 

Prepare  the  fractions  the  same  as  for  Addition,  when 
accessary  ;  then  subtract  the  one  numerator  from  the  other, 
and  set  the  remainder  over  the  common  denominator,  for  the 
diflerence  of  the  fractions  sought. 


EXAMPLES. 

1.  To  find  the  difference  between  |  and  |. 

Here  |  —  e  ~  e  ~  l»  the  Answer. 

2.  To  find  the  difference  between  f  and  |. 
I  —  f  =  If  —  fl  =  A»  the  Answer. 

3.  What 


MULTIPLICATION  OF  VULGAR  FRACTIONS.    63 

3.  What  is  the  difference  between  /^  and  y'^j  ?  Ans.  |. 

4.  What  is  the  difference  between  j\  and  3V  ?        Ans.  ^^j. 
$.  What  is  the  difference  between  f'j  and  x\  ?       Ans.  ^3^. 

6.  What  is  the  diff.  between  5f  and  f  of  4}  ?       Ana.  4/6V- 

7.  What  is  the  difference  between  f  of  a  pound,  and  |  of 
a  of  a  shilling  ?  Ans.  Vt  «  or  I'Os  7d  l^y. 

8.  What  is  the  difference  between  f    of  6^    of  a    pound, 
and  I  of  a  shilling  ?  Ans.  ^^l  or  11  ds  11  Jl^. 


MULTIPLICATION  OF  VULGAR  FRACTIONS. 

*  Reduce  mixed  numbers,  if  there  be  any,  to  equivalent 
fractions  ;  then  multiply  all  the  numerators  together  for  a 
numerator,  and  all  the  denominators  to2;ether  for  a  denomi- 
nator, which  will  give  the  product  required. 


EXAMPLES; 

1.  Required  the  product  of  f  and  f. 

Here  f  X  |  =  /g  =  ^'  ^^^  Answer. 

2.  Required  the  continual  product  of  |,  3^,  5,  and  f  off. 

g        13       ^         3         3         13  X  3       39 

Here  —  X— X— X— X—  = =  —  =  4|  Ans. 

^         4  1         4,2       JS  4X2        8 

3.  Required  the  product  of  f  and  f .  Ans.  -^j. 

4.  Required  the  product  of  y*j  and  /j.  Ans.  j\, 

5.  Required  the  product  of  |,  f ,  and  ||.  Ans.  ^\. 


*  Multiplication  of  any  tking  by  a  fraction,  implies  the  taking  some 
part  or  parts  of  the  thing  ;  it  may  tJierefore  be  truly  expressed  by  a 
compound  fraction  ;  which  is  resolved  by  multiplying  together  the 
numerators  and  denominators. 

Noie^  A  Fraction  is  best  multiplied  by  an  integer,  by  dividing  the 
denominator  by  it ;  but  if  \t  will  not  exactly  divide,  then  multiply  the 
numerator  by  it. 

6.  Required 


|S4  ARITHMETIC. 

6.  Required  the  product  of  i,|,  and  3  Ans.  1. 

7.  Required  the  product  of  i,  |,  and  4/^ .  Ans.  2^V' 

8.  Required  the  product  off,  and  |  of ^.  Ans.  ^f . 

9.  Required  the  product  of  6,  and  f  of  5.  Ans.  20. 
10.  Required  the  product  off  of |,  and  f  of  3f.  Ans.  ||. 
U.  Required  the  product  of  3f  and  4i|  Ans.  14i|f. 
12.  Required  the  product  of  5,  f,  f  off,  and  4^.  Ans.  2-ij. 


DIVISION  OF  VULGAR  FRACTIONS. 


*  Prepare  the  fractions  as  before  in  multiplication  ;  then 
divide  the  numerator  by  the  numerator,  and  the  denominator 
by  the  denominator,  if  they  will  exactly  divide  :  but  if  not, 
then  invert  the  terms  of  the  divisor,  and  multiply  the  dividend 
by  it,  as  in  multiplication. 


EXAMPLES. 

1.  Divide  V  ^7  f  • 

Here  V  -s-  f  ==  f  =  1|,  by  the  first  method. 

2.  Divide!  by f 3. 

Heref  ^  ^^^  :^  f  X  »^  =  f  X  f  =  V  =  4^. 

3.  It  is  required  to  divide  if  by  |.  Ans.  f . 

4.  It  is  required  to  divide  y*^  by  |.  Ans.  y'g. 
6.  It  is  required  to  divide  V  ^J  e  •  •^'^s*  H* 

6.  It  is  required  to  divide  f  by  y .  Ans.  y\. 

7.  It  is  required  to  divide  ||  by  |.  Ans.  ^. 

8.  It  is  required  to  divide  f  by  |.  Ans.  |f . 


*  Division  being  the  reverse  of  Multiplication,  the  reason  of  the 
Rule  is  evident. 

NotCy  A  fraction  is  best  divided  by  an  intecjer,  by  dividing  the  nume- 
rator by  it  ;  bat  if  it  will  not  exactly  divide,  then  multiply  the  deno- 
minator by  it: 

9.  rt 


RULE  OF  THREE  IN  VULGAR  FRACTIONS.       61 

9.  It  is  required  to  divide  tV  l>y  3.  Ans.  f^. 

10.  It  is  required  to  divide  ^  by  2.  Ans.  f^* 

11.  It  is  required  to  divide  7^  by  9f.  .  ,   Ans.  ^f. 

12.  It  is  required  to  divide  f  of  |  by  4  of  7|.  Ans.  j^j. 


RULE  OF  THREE  IN  VULGAR  FRACTIONS. 

Make  the  necessary  preparations  as  before  directed  ;  then 
multiply  continually  together,  the  second  and  the  third  terms, 
and  the  first  with  its  parts  inverted  as  in  Division,  for  th^ 


EXAMPLES. 

1.  If  I  of  a  yard  of  velvet  cost  |  of  a  pound  sterling  ;  what 
will  y\  of  a  yard  cost  ? 

3        2  6  8         jgf         ^. 

—  :  —  :   :  —  :     —  X  —  X  —  =  il  =  6s  Qd,  Answer. 

S        5         16  5         g       XJ& 

2.  What  will  3|  oz  of  silver  cost,  at  6s  4d  an  ounce  ? 

Ans.  1/  1*  4^d. 

3.  If  j\  of  a  ship  be  worth  273^  2*  6d  ;  what  are  /^  of  het 
worth  ?  Ans.  227/  12s  Id, 

4.  What  is  the  purchase  of  1230/  bank-stock,  at  108f  per 
cent.  ?  Ans.  1336/  Is  dd. 

6.  What  is  the  interest  of  273/  15s  for  a  year,  at  3i  per 
cent.  ?  Ans.  8/  17*  ll^d. 

6.  If  I  of  a  ship  be  worth  73/  Is  3d ;  what  part  of  her  is 
worth  250/  10«  ?  Ans.  |. 

7.  What  length  must  be  cut  off  a  board  that  is  7|  inches 
broad,  to  contain  a  square  foot,  or  as  much  as  another  piege 
of  12  inches  long  and  12  broad  ?  Ans.  18i^  inches. 

8.  What  quantity  of  shalloon  that  is  |  of  a  yard  wide,  wiU 
line  9i  yards  of  cloth,  that  is  2^  yards  wide  ?     Ans.  31 J  yds. 


*  This  is  only  multiplying'  the  2d  and  3d  terms  together,  and 
dividing  the  product  by  the  lirst,  as  in  the  Rule  of  Three  in  whole 
nurai>ers. 

Vox..  I.  10  9.  If 


e.G  *  ARITHMETIC. 

9.  If  the  penny  loaf  weighs  6^^  02,  when  the  price  of 
wheat  is  5s  the  bushel ;  what  ought  it  to  weigh  when  the 
wheat  is  8s  6d  the  bushel  ?  Ans.  4  y\  oz. 

10.  How  much  in  length,  of  a  piece  of  land  that  is  11}^ 
poles  broad,  will  make  an  acre  of  land,  or  as  much  as  40 
poles  in  length  and  4  in  breadth  ?  Ads.  ISjVt  poles. 

11.  If  a  courier  perform  a  certain  journey  in  35^  days, 
travelling  13f  hours  a  day;  how  long  would  he  be  in  per- 
forming the  same,  travelling  only  11/^  hours  a  day  ? 

Ans.  40f  jf  days. 

12.  A  regiment  of  soldiers,  consisting  of  976  men,  are  to 
be  new  cloathed  ;  each  coat  to  contain  2i  yards  of  cloth 
that  is  If  yard  wide,  and  lined  with  shalloon  5.  yard  wide  : 
how  many  yards  of  shalloon  will  line  them  ? 

Ans.  4531  yds  1  qr  2^  nails. 


DECIMAL  FRACTIONS. 


A  Decimal  Fraction,  is  that  which  has  for  its  deno- 
minator an  unit  (1),  with  as  many  ciphers  anne^xed  as  the 
numerator  has  places  ;  and  it  is  usually  expressed  by  setting 
down  the  numerator  only,  with  a  point  before  it,  on  the  left- 
band.  Thus,  j%  is  -4,  and  ^-^^  is  -24,  and  j^-^  is  -074,  and 
ToWoo  is  -00124  ;  where  ciphers  are  prefixed  to  make  up  as 
many  places  as  are  ciphers  in  the  denominator,  when  there 
is  a  deficiency  of  figures. 

A  mixed  number  is  made  up  of  a  whole  number  with  some 
decimal  fraction,  the  one  being  separated  from  the  other  by 
a  point.     Thus,  3*25  is  the  same  as  Sy^/^,  or  flf. 

Ciphers  on  the  right-hand  of  decimals  make  no  alteration 
in  their  value  ;  for  -4  or  40,  or  400,  are  decimals  having  all 
the  same  value,  each  being  =  yV'  ^^  I-  ^"^  when  they  arc 
placed  on  the  left-hand  they  decrease  the  value  in  a  ten-fold 
proportion  :  Thus,  -4  is  ^^,  o^  4  tenths  :  but  -04  is  Only  y|^j 
or  4  hundredths,  and  '004  is  only  tAo?  or  4  thousandths. 

The  1st  place  of  decimals,  counted  from  the  left-hand  to- 
wards the  right,  is  called  the  place  of  primes,  or  lOths  ;  the 
2d  is  .the  place  of  seconds,  or  lOOUi  ;  the  3d  is  the  place  of 
thirds,  or  lOOOths  ;  and  so  on.  For.  in  decimals,  as  well  as 
in  whole  numbers,  the  values  of  the  places  increase  towards 
the  left-hand,  and  decrease  towards  the  right,  both  in  the 

same 


ADDITION  OF  DECIMALS.  61 

same  tenfold  proportion  ;  as  in  the  following  Scale  or  Table 
of  Notation. 


1  III  I  I  I  I  \§  I  I  i  t 

33S33S3333333 


.  ADDITION  OF  DECIMALS. 

Set  "the  numbers  under  each  other  according  to  the  value 
«f  their  places,  like  as  in  whole  numbers  ;  in  which  state  the 
decimal  separating  points  will  stand  all  exactly  under  each 
other-  Then  beginning  at  the  right-hand,  add  up  all  the 
columns  of  numbers  as  in  integers  ;  and  point  off  as  many 
places,  for  decimals,  as  are  in  the  greatest  number  of  decimal 
places  in  any  of  the  lines  that  are  added  ;  or  place  the  point 
directly  below  all  the  other  points. 

EXAMPLES.  , 

1.  To  add  together  29-0146,  and  3148-6,  and  2109,  and 
•62417,  and  14-16. 

29-0145 
3146-5 
210^ 

•62417 
14-16 


5299-29877  the  Sum. 


Ex.  2.  What  is    the   sum    of  276,  39-213,  72014-d,  417, 
and  6032  ?  .  \        . 

3.  What  is    the   sum   of  7530,    16-201,  3-014^,  957-13, 
6-72119  and  -03014. 

4.  What  is  the  sum  of  312-09,    3-5711,  7195-6,   71-498, 
9739-216,  179,  and  -0027  ? 

SUBTRACTION 


SS  ARITHMETIC. 


SUBTRACTION  OF  DECIMALS. 

Place  the  numbers  under  each  other  according  to  tJbe 
ualue  of  their  places,  as  in  the  last  Rule.  Then,  beginning 
at  the  right-hand,  subtract,  as  in  whole  numbers,  and  point 
off  the  decimals  as  in  Addition. 


IIXAMPLES, 

t.  To  find  the  difference  between  91-73  and  2-138. 
91-73 
2-138 


Ans.  89-592  the  Difference. 


2.  Find  the  diff.  between  1-9185  and  2-73.  Ans.  0-8116 

3.  To  subtract  4-90142  from  214-81.  Ans.  209-90858, 

4.  Find  the  diff.  between  2714  and  -916.     An§.  2713-084. 


MULTIPLICATION  OF  DECIMALS. 

*  Place  the  factors,  and  multiply  them  together  the  same 
as  if  they  were  whole  numbers. — Then  point  off  in  the  pro» 
duct  just  as  many  places  of  decimals  as  there  are  decimals  in 
both  the  factors.  But  if  there  be  not  so  many  figures  in  the 
product,  then  supply  the  defect  by  prefixing  ciphers. 


*  The  Rule  will  be  evident  from  this  example  : — Let  it  be  re- 
qyiired  to  multiply  '12  by  361 ;  these  numbers  are  eqiiivaknt  to 
J^  and  -3_6_L.  ;  the  product  of  which  is  -j-f  f  f  f  „  =  '04332,  by  the  na- 
ture of  Notation,  which  consists  of  as  many  places  as  there  are  ciphers, 
that  is,  of  as  many  places  as  there  are  in  both  numbers.  And  in  like 
manner  for  any  other  numbers. 


EXAMPLES. 


MULTIPLlCATiON  OF  DECIMALS.  69 


EXAMPLES. 

1.  Multiply  -321096 
by       -2466 

1605480 
1926676 
1284384 
642192 


Ans.  -0791501640  the  Product. 


2.  Multiply  79-347  by  23-15.  Ans.   1836-88306. 

3.  Multiply  -63478  by  -8204.  Ans.  -520773512. 

4.  Multiply  -385746  by  -00464.  Ans.   -00178986144. 

CONTRACTION  I. 

To  multiply  Decimals  by  1  with  any  number  of  Ciphers^  as  by 
10,  or  100,  or  1000,  ^c. 

This  is  done  by  only  removing  the  decimal  point  so  many 
places  farther  to  the  right-hand,  as  there  are  ciphers  in  the 
multiplier  :  and  subjoining  ciphers  if  need  be. 

EXAMPLES. 


1.  The  product  of  51-3  and  1000  is  61300. 

2.  The  product  of  2-714  and  100  is 

3.  The  product  of  -916  and  1000  is 

4.  The  product  of  21-31  and  10000  is 

CONTRACTION  U. 

To  Contract  the  Operation,  so  as  to  retain  only  as  many  Diecimals 
in  the  Product  as  may  be  thought  Necessary^  when  the  Pro- 
duct would  naturally  contain  several  more  Places. 

Set  the  units'  place  of  the  multiplier  under  that  figure  of 
t-he  multiplicand  whose  place  is  the  same  as  is  to  be  retained 
for  the  last  in  the  product ;  and  dispose  of  the  rest  of  the 
figures  in  the  inverted  or  contrary  order  to  what  they  are 
usually  placed  in. — Then,  in  multiplying,  reject  all  the  figures 
that  are  more  to  the  right-hand  than  each  multiplying  figure, 
and  set  down  the  products,   so  that  their  right-hand  figures 

may 


20  ARlTHMETie.  r.ttiT  I  r 

may  fall  io  a  column  straight  below  each  other  ;  but  observing 
to  increase  the  first  figure  df  every  line  with  what  would 
•arise  from  the  figures  omitted,  in  this  manner,  namely  1  from 
5  to  14,  2  from  15  to  24,  3  from  25  to  34,  &c. ;  and  the  sum 
of  all  the  lines  will  be  the  product  as  required,  commonly  to 
the  nearest  unit  in  the  last  figure. 

EXAMPLES. 

1.  To  multiply  27»14986  by  92-41036,  so  as  to  retain  oply 
four  places  of  decimals  in  the  product. 

Contracted  Way.  Common  Way. 
72-14986  27  14S86 

530 14 -29  92-41035 


24434874 

542997 

108599 

2715 

81 

14 

2508-9280 


13 

574930 

81 

44958 

2714 

986 

103599 

44  V 

642997 

2 

24434874 

2608-9280 

650510 

2.  Multiply  480-14936  by  2-72416,   retaining  only  four  de- 
cimals in  the  product. 

3.  Multiply  2490-3048  by  -573286,  retaining  only  five  de- 
cimals in  the  product. 

4.  Multiply  325-701428  by  -7218393,  retaining  only  three 
^lecimals  in  the  product. 


DIVISION  OF  DECIMALS. 

Divide  as  in  whole  numbfers  ;  and  point  off  in  the  quotient 
as  many  places  for  decimals,  as  the  decimal  places  in  the  divi- 
dend exceed  those  in  the  divisor*. 


♦  The  reason  of  this  Rule  is  evident ;  for,  since  the  divisor  multiplied 
by  the  quotient  gives  the  dividend,  therefore  the  number  of  decimal 
places  in  the  dividend,  is  equal  to  those  in  the  divisor  and  quotient, 
taken  together,  by  the  nature  of  Multiplication  ;  and  consequently  the 
qs>''tient  itself  must  contain  as  many  as  the  dividend  exceeds  the  di- 
^or- 

Another 


DIVISION  OF  DECIMALS. 


71 


Another  way  to  know  the  .place  for  the  decimal  point,  is* 
this  :  The  first  figure  of  the  quotient  must  be  made  to  oc- 
cupy the  same  place,  of  integers  or  decimals,  as  doth  that 
figure  of  the  dividend  which  stands  over  the  unit's  figure  of  the 
the  first  product. 

When  the  places  of  the  quotient  are  not  so  many  as  the 
Rule  requires,  the  defect  is  to  be  supplied  by  prefixing 
ciphers. 

When  there  happens  to  be  a  remainder  after  the  division, 
or  when  the  decimal  places  in  the  divisor  are  more  than  those 
in  the  dividend  ;  then  ciphers  may  be  annexed  to  the  dividend, 
and  the  quotient  carried  on  as  far  as  required. 

EXAMPLES. 


1. 
178)  -48520998  (-00372589 
1292  . 
460 
1049 
1599 
1768 
166 


•2639)  27-00000  (  102-3114 
6100 
8220 
303U 
3910 
12710 
2154 


3.  Divide  123-70536  by  64-25. 

4.  Divide  12  by  7854. 

5.  Divide  4195-68  by  100. 

6.  Divide  -8297592  by  -153. 


Ans.  2-2802. 

Ans.  15-278. 

Ans.  41-9568. 

Ans.  5-4232. 


CONTRACTION  1. 


When  the  divisor  is  an  integer,  with  any  number  of  ciphers 
annexed  :  cut  off  those  ciphers,  and  remove  the  decimal 
point  in  the  dividend  as  many  places  farther  to  the  left  as 
there  are  ciphers  cut  off,  prefixing  ciphers  if  need  be  :  then 
proceed  as  before.* 


*  This  is  no  more  than  dividing  both  divisor  and  dividend  by  the 
same  number,  either  10,  or  100,  or  1000,  &c.  according  to  the  number 
of  ciphers  cut  off,  which  leaving  them  in  the  same  proportion,  does 
not  affect  the  quotient.  And,  in  the  same  way,  the  decimal  point  may 
be  moved  the  same  number  of  places  in  both  the  divisor  and  dividend, 
either  to  the  right  or  left,  whether  they  have  ciphers  or  not. 


EXAMPLES. 


7g  ARITHMETIC. 

EXAMPLES^ 

1.  Divide  46-5  by  2100. 

21-00  )  -455  (  -0216,  mi. 
35 
140 
14 

2.  DiTide  41020  by  32000. 

3.  Divide       953  by  21600. 

4.  Divide         61  by  79000. 

CONTRACTION  It 

Hence,  if  the  divisor  be  1  with  ciphers,  as  10,  K)0,  oj 
1000,  &c.  :  then  the  quotient  will  be  found  by  merely  mov- 
ing the  decimal  point  in  the  dividend,  so  many  places  farther 
to  the  left,  as  the  divisor  has  ciphers  ;  prefixing  ciphers  if 
need  be. 

EXAMPLES. 

So,  217-3 -r  100  =  2-173  And  419  ~     10  = 

And  6-16  -^  100  =  And  21  -f-  1000  = 

CONTRACTION  HI. 

When  there  are  many  figures  in  the  divisor  ;  or  when  only 
a  certain  number  of  decimals  are  necessary  to  be  retained  ia 
the  quotient :  then  take  only  as  many  figures  of  the  divisor 
as  will  be  equal  to  the  number  of  figures,  both  integers  and 
decimals,  to  be  in  the  quotient,  and  find  how  many  times 
they  may  be  contained  in  the  first  figures  of  the  dividend,  as 
usual. 

Let  each  remainder  be  a  new  dividend  ;  and  for  every  such 
dividend,  leave  out  one  figure  more  on  the  right-hand  side  of 
the  divisor  ;  remembering  to  carry  for  the  increase  of  the 
figures  cut  off,  as  in  the  2d  contraction  in  Multiplication. 

JVote.  When  there  are  not  so  many  figures  in  the  divisor,  as 
are  required  to  be  in  the  quotient,  begin  the  operation  with  all 
the  figures,  and  continue  it  as  usual  till  the  number  of  figures 
in  the  divisor  be  equal  to  those  remaining  to  be  found  in  the 
quotient :  after  which  begin  the  contraction. 

EXAMPLES. 

1.  Divide  2508-92806  by  92-41035,  so  as  to  have  only  four 
decimals  in  the  quotient,  in  whick  case  the  quotient  will  con- 
tain six  figures. 

Contracted. 


REDUCTION  OF  DECIMALS.  73 

Contracted.  Common, 


92-41Q3,5)2508-928,06(27- 1498 
660721 
13849 
4608 

80 
'      6 


92-4103,5)2508-928  06(2ri'498 
66072106 
13848610 
46075760 
91116100 
79467850 
5539570 


2.  DivicJe  4109-2351  .by  230-409,  so  that  the  quotient  may 
contain  only  fooT  decimals.  Aas.  17-8346. 

3.  Divide'  37-10438  by  5713-96,  that  the  quotient  may  con- 
tail)  only  live  rfecimalap.  Ans.  -00649. 

4.  Divide  913^08  by  2137-2^  that  the  qijotient  may  contain 
•nly  three  decimals. 


REDUCTION  OF  DECIMALS. 

CASE  I. 

To  reduce  a  Vulgar  Fraction  to  its  equivalent  Decimal, 

Divide  the  numerator  by  the  denominator  as  in  Division 
of  Decimals,  annexing  ciphers  to  the  numerator  as  far  as  ne- 
cessary ,-  so  »half  the  qWtient  be  the  decimal  required. 


1.  Reduce  ^\  to  at  ^cimal. 
24  =  4  X  6.     Then  4)  7- 

6)  1  •750000. 
•29l6b6  &c. 


t.  Reduce  i,  and  ^,  and  |,  to  decimals. 

Ans.  't5,  afwf  '5,  arid  •'^5. 

3.  Redace  f  to  a  decimal.  Ans.  -625. 

4.  Reduce  2  J  to  a  decimal.  Ans.  12. 
6.  Reduce  yfj  toadeciaiaL  Ans  Oj  1350. 
e^.  Reduce  ^/Jv  to  a  decimal.  Ans.  -143155  &c. 

Voc.  I.                                      U  CASE 


74  ARITHMETIC. 


CASE  II. 

To  find  the  Value  of  a  Decimal  in  terms  of  the  Inferior  Deno- 
minations. 

Multiply  the  decimal  by  the  number  of  parts  in  the 
next  lower  denomination  ;  and  cut  off  as  many  places  for  a 
remainder  to  the  right  hand,  as  there  are  places  in  the  given 
decimal. 

Mutiply  that  remainder  by  the  parts  in  the  next  lower 
denomination  again,  cutting  off  for  another  remainder  as 
before. 

Proceed  in  the  same  manner  through  all  the  parts  of  the 
integer  ;  then  the  several  denominations  separated  on  the  left- 
hand,  will  make  up  the  answer. 

JVofe,  This  operation  is  the  same  as  Reduction  Descending 
in  whole  Rumbers. 


EXAMPLES. 

i.  Required  to  find  the  value  of  '775  pounds  sterling. 

•776 
20 


s  15-600 
12 


d  6-000  Ans.  16«  6d, 


2.  What  is  the  value  of  -625  shil  ?  Ans.  7idf. 

3.  What  is  the  value  of  'S63ol?  Ans.  17s  3-24d. 

4.  What  is  the  value  of  -0125  lb  troy  ?  Ans.  3  dwts. 
6.  What  is  the  value  of  -4694  lb  troy  ? 

Ans.  6  oz  12  dwts  15-744  gr. 

6.  What  is  the  value  of  -626  cwt  ?  Ans.  2  qr  14  lb. 

7.  What  is  the  value  of  -009943  miles  ? 

Ans.  17  yd  1  ft  5-98848  inc. 

8.  What  is  the  value  of   6875  yd  ?  Ans.  2  qr  3  nls. 

9.  What  is  the  value  of  -3375  acr  ?        Ans.  1  rd  14  poles. 
10.  What  is  the  value  of  -S^'S  hhd  of  wine  ? 

Ans.  13-1229  gal. 

CASE 


REDUCTION  OF  DECIMALS.  75 

CASE  ni. 

To  reduce  Integers    or   Decimals   to   Equivalent  Decimals  of 
Higher  Denominations* 

Divide  by  the  number  of  parts  in  the  next  higher  deno- 
minatioQ  ;  continuing  the  operation  to  as  many  higher  deno- 
minations as  may  be  necessary,  the  same  as  in  Reduction 
Ascending  of  whole  numbers. 

EXAMPLES. 
1.  Reduce  1  dwt  to  the  decimal  of  a  pound  troy. 


20 
12 


1  dwt 

0-06  oz 

0004166  &c.  lb.  Ans. 


2.  Reduce  9d  to  the  decimal  of  a  pound.         Ans.  '03151. 

3.  Reduce  7  drams  to  the  decimal  of  a  pound  avoird. 

Ans.   •027343751b. 

4.  Reduce  26  J  to  the  decimal  of  a  Z.    Ans.  -0010833  &c. /. 

5.  Reduce  2- 15  lb  to  the  decimal  of  cwt. 

Ans.  -oiaise+cwt. 

6.  Reduce  24  yards  to  the  decimal  of  a  mile. 

Ans.  -013636  &c.  mile. 

7.  Reduce  '056  pole  to  the  decimal  of  an  acre. 

Ans.  -00035  ac. 
€.  Reduce  1*2  pint  of  *vine  to  the  decimal  of  a  hhd. 

Ans.    00238-}-hhd. 
9.  Reduce  14  minutes  to  the  decimal  of  a  day. 

Ans.  -009722  &c.  da. 

10.  Reduce  -21  pint  to  the  decimal  of  a  peck. 

Ans.  -013125  pec. 

11.  Reduce  28"  12'"  to  the  decimal  of  a  minute. 

Note,  When  there  are  several  numbers ^  to  be  reduced  all  to 
to  the  decimal  of  the  highest : 

Set  the  given  numbers  directly  under  each  other,  for  divi^ 
dends,  proceeding  orderly  from  the  lowest  denomination  to 
the  highest. 

Opposite  to  each  dividend,  on  the  left-hand,  set  such  a 
number  for  a  divisor  as  will  bring  it  to  the  next  higher  name  ; 
drawing  a  perpendicular  line  between  all  the  divisors  and 
dividends. 

Begin  at  the  uppermost,  and  perform  all  the  divisions  : 
©nly  observing  to  set  the  quotient  of  each  division,  as  decimal 

parts, 


76  ARITHMETIC. 

parts,  on  the  right-han.l    of  the.  dividpnd  next  below  it 
shall  the  last  quotient  be  the  decimal  required. 


EXAMPl^S. 


1.  Reduce  17s  9fc?  to  the  decimal  of  a  pound. 


4 
12 

20 


3- 

9-75 
17-8125 
i:  ()-89U625  Ans. 


2.  Reduce  ilH  17^  a^^/ to  L  Ans.  19-86354166  kc.  L 

3.  Reduce  15.9  Hrf  to  the  decimal  of  a  L  Ans.   -775^ 

4.  Kedutu^  7.',c/to  the  decimal  of  a  shilHng.  Ans.   '6^55. 

5.  Reduce  5  oz  12  dvvts  16  gr  to  lb.         Ans.  -46044  kc.  lb 


RULE  OF  THREE  IN  DECIMALS. 

Prepare  the  terms  by  reducing  the  vulgar  fractions  to 
decimals,  and  any  compound  numbers  either  to  decimals  of  the 
hii:her  dcnominationsri.  or  to  integers  of  the  lower,  also  the 
first  and  third  terms  to  the  same  name  :  Then  multiply  and 
divide  as  in  whole  numbers. 

J\'ote,  Any  of  the  convenient  Examples  in  the  Rule  of 
Three  or  Rule  of  Five  in  Integers,  or  Vulgar  Fractions,  may 
be  taken  as  proper  examples  to  the  same  rules  in  Decimals. 
' — The  following  Example,  which  is  the  first  in  Vulgar  Frac- 
tions, is  wrought  out  iiere,  to  show  the  method. 

Iff  of  a  yard  of  velvet  cost  f/,  what  will  -fj  yd  cost  ? 

yd        /  yd         /  s  d 

I  =  -375  -376  :  -4  :  :  -3125  :  -333  &c.  or  6  8 

•4 


•375)  -12500     (-333333  &c, 
1250  20 


125 


s  6-66666  &c, 

-^^  —  'snr^  12 


Ans.  6s  8d.  d  7-99999  &c.=8a^ 

DUOIDE. 


DUODECIMALS.  77 


DUODECIMALS. 


Duodecimals  or  Cross  Multiplication,  is  a  rule  used 
by  workmen  and  artificers,  in  computing  the  contents  of  their 
works. 

Dimensions  are  usually  taken  in  feet,  inches,  and  quarters  ; 
any  parts  smaller  than  these  being  neglected  as  of  no  conse- 
quence. And  the  same  in  multiplying  them  together,  or  cast- 
ing up  the  contents.     The  method  is  as  follows. 

Set  down  the  two  dimensions  to  be  multiplied  together,  one 
under  the  other,  so  that  feet  may  stand  under  feet,  inches  un- 
der inches,  &c. 

Multiply  each  term  in  the  multiplicand,  beginning  at  the 
lowest,  by  the  feet  in  the  multiplier,  and  set  the  result  of  each 
straight  under  its  corresponding  term,  observing  to  carry  1  for 
every  1  ^  from   he  inches  to  the  feet. 

In  like  manner,  multiply  all  the  multiplicand  by  the  inches 
and  parts  of  the  multiplier,  and  set  the  result  of  each  term 
one  place  removed  to  the  right-hand  of  those  in  the  multipli- 
cand ;  omitting,  however,  what  is  below  parts  of  inches,  only 
carrying  to  these  the  proper  number  of  units  from  the  lowest 
denomination. 

Or,  instead  of  multiplying  by  the  inches,  take  such  parts  ol 
the  multiplicand  as  there  are  of  a  foot.        • 

Then  add  the  two  lines  together  after  the  manner  of  [Com- 
pound Addition,  carr)'ing  l.to  the  feet  "for  12  inches,  when 
these  come  to  so  many. 


EXAMPLES. 

Multiply  4  f  7  inc              2.  Multiply  14  f  9  inc 
by  6     4                                      by  4     6 

27     6 
1     H 

59     0 
7     41 

Ans.  29     Oi 

Ans.  66     4-L 

3.  Multiply  4  feet  7  inches  by  9  f  6  inc.  Ans.     43  f  6^  inc. 

4.  Multiply  12  f  6  inc  by  6  f  8  inc.  Ans.     82     9i 

5.  Multiply  35  fU  inc  by  12  f  3  inc.  Ans.  433     4} 

6.  Multiply  64  f  6  inc  by  8  f  9^  inc.  Ans.  665     8f 

INVOLUTION. 


71 


ARITHMETIC* 


INVOLUTION. 


InvolVtiok  is  the  raising  of  Powers  from  any  giren  number^ 
as  a  root. 

A  Power  is  a  quantity  produced  by  multiplying  any  given 
number,  called  the  Root,  a  certain  number  of  times  continual- 
ly by  itself.     Thus, 

2  =    2  is  the  root,  or  1st  power  of  2. 
2X2=    4  is  the  2d  power,  or  square  of  2. 
2X2X2=    8  is  the  3d  power,  or  rube  of  2. 
2  X  2  X  2  X  2  =  16  is  the  4th  power  of  2,  &c. 

And  in  this  manner  may  be  calculated  the  following  Table  of 
of  the  first  nine  powers  of  the  first  9  numbers. 

TABLE  OF  THE  FIRST  NINE  POWERS  OF  NUMBERS. 


1st 

1 

2 

2d 
1 
4 

3d 

4th 

5th 

6th 

7th 

8th 

9th 

8 

1 

16 

1 

1 

1 

1 

1 

32 

64 

128 

256 

,612 

3 

9 

27 

81 

243 

729 

2187 

6661 

19683 

4 

16 

64 

256 

1024 

4096 

16384 

65536 

262144 

5 

25 

125 

626 

31  £5 

15625 

78125 

390625 

1953126 

•6 

36 

216 

1296 

7776 

466S6 

279936 

1679616 

10077696 

7 

49 

343 

2401 

168U7 

117649 

823543 

5764801 

40353607 

8 

64 
81 

512 

4096 

32768 

262144 

2097152 

16777216 

134217728 

9 

729 

6561 

59049 

531441 

4782969 

43046721 

387420489 

The 


INVOLUTION.  19 

The  Index  or  Exponent  of  a  Power,  is  the  number  de- 
noting the  height  or  degree  of  that  power  ;  and  it  is  1  more 
than  the  number  of  multiplications  used  in  producing  the 
same.  So  1  is  the  index  or  exponent  of  the  first  power  or 
root,  two  of  the  2d  power  or  square,  3  of  the  third  power  or 
cube,  4  of  the  4th  power,  and  so  on. 

Powers,  that  are  to  be  raised,  are  usually  denoted  by  placing 
the  index  above  the  root  or  first  power. 

So  23  =  4  is  the  2d  power  of  2. 
2»  =  8  is  the  3d  power  of  2. 
2*  =16  is  the  4th  power  of  2, 
540*       is  the  4tb  power  of  640,  &c. 


When  two  or  more  powers  are  multiplied  together,  their 
product  is  that  power  whose  index  is  the  sum  of  the  expo- 
nents of  the  factors  or  powers  multiplied.  Or  the  multipli- 
cation of  the  powers,  answers  to  the  addition  of  the  indices. 
Thus,  in  the  following  powers  of  2, 

1st     2d     3d     4th     6th     6th     7th     8th     9th     10th 
2      4       8        IG       32       64      128     256     512   1024 
or  2»    22     2»     2*       2*       2^      2^      2^       2*     2»» 

Here,  4  X  4  =  16,  and  2  -f  2  =  4  its  index  ; 
and  8  X  16  =  128,  and  3  -f  4  =  7  its  index  ; 
also  16  X  64  =     1024,  and  4  -f-  6  =  10  its  index. 


OTHER  EXAMPLES. 

1.  What  is  the  2d  power  of  45  ?  Ans.  2025. 

2.  What  is  the  square  of  4-16  ?  Ans.  17-3056. 

3.  What  is  the  3d  power  of  3-5  ?  Ans.  42-875. 

4.  What  is  the  6th  power  of  -029  ?  Ans.  -00000002051 1 149. 

5.  What  is  the  square  of  |  ?  Ans.  f . 

6.  What  is  the  3d  power  of  f  ?  Ans.  iff. 

7.  What  is  the  4th  power  of  f  ?  Ans.  /jV- 


JSVOLUTION. 


80  ARITHMETIC. 


EVOLUTION. 

EvottfnoN,  or  the  reverse  of  Involution,  is  the  extracting 
or  finding  the  roots  of  any  given  powers. 

The  root  of  any  number,  or  power,  is  such  a  number,  as 
being  multiplied  into  itself  a  certain  number  of  times,  will 
produce  that  power.  Thus,  2  is  the  square  root  or  2d  root 
of  4,  because  23  =  2  X  2  =  4  ;  and  3  is  the  cube  root  or  3d 
root  of  27,  because  S^  =  3  X  3  X  3  ~  27. 

Any  power  of  a  given  number  or  root  may  be  found  ex- 
actly, namely,  by  multiplying  the  number  continually  into 
itself.  But  there  are  many  numbers  of  which  a  proposed  root 
can  never  be  exactly  found.  Yet,  by  means  of  decimals,  we 
may  approximate  or  approach  towards  the  root,  to  any  de^ 
gr6e  of  exactness. 

Those  roots  which  only  approximate,  are  called  Surd 
roots  ;  but  those  which  can  be  found  quite  exact,  are  called 
Rational  Roots.  Thus,  the  square  root  of  3  is  a  surd  root  ; 
but  the  square  root  of  4  is  a  rational  root,  being  equal  to  2  : 
also  the  cube  root  of  8  is  rational,  being  eq.ual  to  2  ;  but  the 
cube  root  of  9  is  surd  or  irrational. 

Roots  are  sometimes  denoted  by  writing  the  character  ^ 
before  the  power,  with  the  index  of  the  root  against  it. 
Thus,  the  3d  root  of  20  is  expressed  by  ^  20  ;  and  the  square 
root  or  2d  root  of  it  is  ^  20,  the  index  2  being  always  omit- 
ted, when  only  the  square  root  is  designed. 

When  the  power  is  expressed  by  several  numbers,  with  the 
sign  -h  or  —  between  them,  a  line  is  drawn  from  the  top  of 
the  sign  over  all  the  parts  of  it  :  thus  the  third  root  of 
45  —  12  is  ^  45—  12,  or  thus  ^  (45  --.  12),  inclosing  the 
numbers  in  parentheses. 

But  all  roots  are  now  often  designed  like  powers,  with 
fractional  indices  :  thus,  the  square  root  of  8  is  8^,  the  cube 
root  of  25  is  253,  and  the  4th  root  of  45  —  18 is  45— 18)*, 
or  (45— 18)^. 


SQUARE  ROOT.  81. 


TO  EXTRACT  THE  SQUARE  ROOT. 

*  Divide  the  given  number  into  periods  of  two  figures 
each,  by  setting  a  point  over  the  place  of  units,  another  over 
the  place  of  hundreds,  and  so  on,  over  every  second  figure, 
both  to  the  left-hand  in  integers,  and  te  the  right  in  deci- 
mals. 

Find  the  greatest  square  in  the  first  period  on  the  left-hand, 
and  set  its  root  on  the  right  hand  of  the  given  number,  after 
the  manner  of  a  quotient  figure  in  Division. 


*  The  reason  for  separating  the  figures  of  the  dividend  into  periods 
or  portions  of  two  places  each,  is,  that  the  square  of  any  single  figure 
never  consists  of  more  than  two  places  ;  the  square  of  a  number  of 
two  figures,  of  not  more  than  four  places,  and  so  on.  So  that  there 
will  be  as  many  figures  in  the  root  as  the  given  number  contains  periods 
so  divided  or  parted  off. 

And  the  reason  of  the  several  steps  in  the  operation  appears  from 
the  algebraic  form  of  the  square  of  any  number  of  terms,  whether  two 
or  three  or  more.     Thus, 

(a  +  ^)^=a«  +  2ab  +  b*=a^  +  (2a  +  b)  6,  the  square  of  two 
terms  ;  where  it  appears  that  a  is  the  first  term  of  the  root,  and  b 
the  second  term  j  also  a  the  first  divisor,  and  the  new  divisor  is 
2a  +  by  or  double  the  first  term  increased  by  the  second.  And  hence 
the  manner  of  extraction  is  thus  : 

.    1st  divisor  a)  az  -j- 2ab  +  b^  (  a  +  6  the  root. 

2d  divisor  2a  +  6  I  2(2^  +  b- 
b\2ab  ~^b2 


Again,  for  a  root  of  three  parts,  a,  b,  c,  thus  : 

(a  +  6  -f-  c)  2       fl2  +  2a6  -f-  62  -f-  2ac  -|-  2bc  +  c^  = 

a2  +  {2a  +  6)  A  +  (2a  4  26  +  c)  c,  the 
square  of  three  terms,  where  a  is  the  first  term  of  the  root  b,  the 
second,  and  c  the  third  term  ;  also  a  the  first  divisor,  2a  ^b  the 
second,  and  2a  ■\'2b  -f-  c  the  third,  each  consisting  of  the  double  of 
the  root  increased  by  the  next  term  of  the  same.  And  the  mode  of 
extraction  is  thus  : 
1st  divisor  a)  a^  -{•  2ab  +  b"^  +  2ac-f-2dc  -f-  c'  (a  +  6  -f-  c  the  root. 


2d  divisor  2a'^b\2ab  +  b^ 
b\2ab-\-  b^ 


od  divisor  2<i  f  26  -f  c\  2ac  -f  2bc  +  c* 
c  I  2ac  +  26c  +  c2 

Vol.  I.  }%^  Subtract 


82  ARlTHMEtie. 

Subtract  the  square  thus  found  from  the  said  period,  atid 
to  the  remaiiidar  annex  the  two  figures  of  the  next  following 
period,  for  a  dividend. 

Double  the  root  above  mentioned  for  a  divisor  ;  nnd  find 
how  often  it  is  contained  in  the  said  dividend,  exclusive  of 
its  right-hand  figure  ;  and  set  that  quotient  figure  both  in  the 
quotient  and  divisor. 

Multiply  the  whole  augmented  divisor  by  this  last  quotient 
figure,  and  subtract  the  product  from  the  said  dividend, 
bringing  down  to  it  the  next  period  of  the  given  number,  for 
a  new  dividend. 

Repeat  the  same  process  over  again,  viz.  find  another  new 
divisor,  by  doubling  all  the  figures  now  found  in  the  root ; 
from  which,  and  the  last  dividend,  find  the  next  figure  of 
the  root  as  before  ;  and  so  on  through  all  the  periods,  to  the. 
last. 

Note,  The  best  way  of  doubling  the  root,  to  form  the  new 
divisors,  is  by  adding  the  last  figure  always  to  the  last  divisor, 
as  appears  in  the  following  examples. — Also,  after  the  figures 
belonging  to  the  given  number  are  all  exhausted,  the  opera- 
tion may  be  continued  into  decimals  at  pleasure,  by  adding  any 
number  of  periods  of  ciphers,  two  in  each  period. 

EXAMPLES. 

1.  To  find  the  square  root  of  29506624. 

29506624  (  6432  the  root; 
25 


104 

4 

450 
416 

1083 
3 

34G6 
3249 

10862 

2 

21724 
21724 

Note,  When  the  root  is  to  he  extracted  to  many  places  of  figures  ^ 
the  work  may  he  considerably  shortened^  thus  • 

Having  proceeded  in  the  extraction  after  the  common  me- 
fliod,  till  there  be  found  half  the  required  number  of  figures 


SQ.UARE  ROOT. 


8S 


in  the  root,  or  one  figure  more  ;  then,  for  the  rest,  divide 
the  last  remainder  by  its  corresponding  diWsor,  after  the  man- 
ner of  the  third  contraction  in  Division  of  Decimals  ;  thus, 
2.  To  find  the  root  of  2  to  nine  places  of  figures. 

2  (  1 '4 142 1366  the  root. 

1 


24 
4 

100 
96 

281 

1 

400 
281 

2824   11900 
4   11296 

28282 
2 

60400 

6G5G4 

28284  ) 


3836  ( 
1008 
160 
19 


1356 


3.  What 

4.  What 

5.  What 

6.  What 

7.  What 

8.  What 

9.  What 
10.  What 
Ih  What 
12.  What 


is  the  square  root  of  2025  ? 
is  the  square  root  of  17-3056  ? 
is  the  square  root  of  -000729  ? 
is  the  square  root  of  3  ? 
is  the  square  root  of  5  ? 
is  the  square  root  of  6  ? 
is  the  square  root  of  7  ? 
is  the  square  root  of  10  ? 
is  the  square  root  of  11  ? ' 
is  the  square  root  of  12  ? 


Ans.  4S. 

Ans.  416. 

Ans.  -027. 
Ans.  1-732050. 
Ans.  2-236068. 
Ans.  2-449489. 
Ans.  2-643751. 
Ans.  3-162277. 
Ans.  3-316624. 
Ans.  3-464101. 


RULES  FOR  THE  SQUARE   HOOTS  OF  VULGAR  FRACTIONS 
AND  MIXED  NUMBERS. 


First  prepare  all  vulgar  fractions,  by  reducing  them  to 
their  least  terms,  both  for  this  and  all  other  roots.     Then 

1.  Take  the  root  of  the  numerator  and  of  the  denominator 
for  the  respective  terms  of  the  root  required.  And  this  is  the 
best  Way  if  the  denominator  be  a  complete  power  :  but  if  it 
be  not,  then 

2.  Multiply  the  numerator  and  denominator  together  ; 
take  the  root  of  the  product :  this  root  being  made  the  nume 

rator 


84 


ARITHMETIC. 


rator  to  the  denominator  of  the  given  fraction,  or  made  the 
denominator  to  the  numerator  of  it,  will  form  the  fractional 
root  required. 


Thatis,y---  --^^ 


^ab. 


And  this  rule  will  serve,  whether  the  root  be  finite  or  infinite. 

3.  Or  reduce  the  vulgar  fraction  to  a  decimal,  and  extract 
its  root. 

4.  Mixed  numbers  may  be  either  reduced  to  improper 
fractions,  and  extracted  by  the  first  or  second  rule,  or  the 
vulgar  fraction  may  be  reduced  to  a  decimal,  then  joined  to 
tUe  integer,  and  the  root  of  the  whole  extracted. 

EXAMPLES. 


i.  What  is  the  root  of  |f  ?  Ans.  -f . 

2.  What  is  the  root  of  ^Vt  ?  Ans.  -f. 

3.  What  is  the.  root  of  yV  ?  Ans.  0-866025. 

4.  What  is  the  root  of  -f^  1  Ans.  0-645497. 

5.  What  is  the  root  of  17f  ?         '  Ans.  4-168333. 
By  means  of  the  square  root  also  may  readily  be  found  the 

4th  root,  or  the  8th  root,  or  the  16th  root,  &c.  tlrat  is,  the 
root  of  any  power  whose  index  is  some  power  of  the  number 
2  ;  namely,  by  extracting  so  often  the  square  root  a»s  is  de- 
noted by  that  power  of  2  :  that  is,  two  extractions  for  the 
4th  root,  three  for  the  8th  root,  and  so  on. 

So,  to  find  the  4th  root  of  the  number  21035'8,  extract 
the  square  root  two  times  as  follows  z 


21035  8000 
1 


(145  037237  (12-0431407  the  4th  root. 

X 


24  I  110 
4   96 


22  I  45 
21  44 


285  I 

6  ' 


1435 
1425 


2404 

4 


10372 
9616 


29003 
3 


108000  24083 
87009     3 


75637 

72249 


,  20991   (.7237         3388  (1407 
687  980 

107  17 

Ex.  2.  What  is  the  4th  root  of  97-41  ? 


To 


CUBE  RO0T.  B5 

TO  EXTRACT  THE  CUBE  ROOT. 
1.  By  the  Common  Rule,* 

1.  Having  divided  the  given  number  into  periods  of  three 
figures  each,  (by  setting  a  point  over  the  place  of  units,  and 
also  over  every  third  figure,  from  thence,  to  the  left  hand  in 
whole  numbers,  and  to  the  right  in  decimals),  find  the  nearest 
less  cube  to  the  first  period  ;  set  its  root  in  the  quotient,  and 
subtract  the  said  cube  from  the  first  period  ;  to  the  remainder 
brine  down  the  second  period,  and  call  this  the  resolvend. 

2.  To  three  times  the  square  of  the  root,  just  found,  add 
three  times  the  root  itself,  setting  this  one  place  more  to  the 
right  than  the  former,  and  call  this  sum  the  divisor.  Then 
divide  the  resolvend,  wanting  the  last  figure,  by  the  divisor, 
for  the  next  figure  of  the  root,  which  annex  to  the  former ; 
calling  this  last  figure  e,  and  the  part  of  the  root  before  found 
let  be  called  a. 

Add  all  together  these  three  products,  namely,  thrice  a 
square  multiplied  by  e,  thrice  a  multiplied  by  e  square,  and 
e  cube,  setting  each  of  them  one  place  more  to  the  right  than 
the  former,  and  call  the  sum  the  subtrahend  ;  which  must 
not  exceed  the  resolvend  ;  but  if  it  does,  then  make  the  last 
figure  e  less,  and  repeat  the  operation  for  finding  the  subtra- 
hend, till  it  be  less  than  the  resolvend. 

4.  From  the  resolvend  take  the  subtrahend,  and  to  the  re- 
mainder join  the  next  period  of  the  given  number  for  a  new 
resolvend  ;  to  which  form  a  new  divisor  from  the  whole  root 
now  found  ;  and  from  thence  another  figure  of  the  root,  as 
directed  in  Article  2,  and  so  on. 


*  The  reason  for  pointing  the  given  number  into  periods  of  three 
figures  each,  is  because  the  cube  of  one  figure  never  amounts  to  more 
than  three  places.  And,  for  a  siinilar  reason,  a  given  number  is  point- 
ed into  periods  of  four  figures  for  the  4th  root,  of  five  figures  for 
the  5tli  root,  and  so  on.  • 

And  the  reason  for  the  other  parts  of  the  rule  depends  on  the 
algebraic  formation  of  a  cube  :  for  if  the  root  consists  of  the  two 
parts  a  +  /;,  then  its  cube  is  as  foUow^s  :  (a  +  6)^  =a3  -J-  3a2b-j- 
3(ibs  +  ^3  ;  where  a  is  the  root  of  the  first  part  u^  ;  the  resolvend  is 
Sa^b  •{•  5ab2  -f- ^^j  which  is  also  the  same  as  the  three  parts  of 
the  subtrahend  ;  also  the  divisor  is  3v3  -|-  3a,  by  which  dividing  the 
first  two  terms  of  the  resolvend  3a^  b  +  ab^ ,  gives  b  for  the  second 
part  of  the  root ;  and  so  on. 

EXAMPLE. 


96  ARITHMETIC. 

EXAMPLE. 

To  extract  the  cube  root  of  48228-544. 


3  X  32  =  27 
3X3    =    09 

Divisor  279 


48228-544  (36-4  root. 
27 


21228  resolvend. 


3  X  32   X  6    =162      ) 
3X3     X  62  =    324    >  add 


63=      216' 


3  X  362  =  3888 
3  X  36    =       108 


38988 


19666  subtrahend. 


1572544  resolvend. 


3  X  362   X  4    =  15552 
3  X  36     X  42  =       1728)  add 
43  =  64  ^ 


1572544  subtrahend. 


0000000  remainder. 


Ex.  2.  Extract  the  cube  root  of  571482-19. 
Ex.  3.  Extract  the  cube  root  of  1628-1582. 
Ex.  4.  Extract  the  cube  root  of  1332. 


II.  To  extract  the  Cube  Root  by  a  short  Way.* 

1.  By  trials,  or  by  the  table  of  roots  at  p.  90,  &c.  take  the 
nearest  rational  cube  to  the  given  number,  whether  it  be 
greater  or  less  ;  and  call  it  the  assumed  cube. 

2.  Then 


*  The  method  usually  given  for  extracting  the  cube  root,  is  so 
exceedingly  tedious,  and  difficult  to  be  remembered,  that  various 
other  approximating  rules  have  been  invented;  viz.  by  Newton, 
Raphson,  Halley,  De  Lagny,  Simpson,  Emerson,  and  several  other 
mathematicians  ;  but  no  one  that  I  have  yet  seen,  is  so  simple  in 
its  form,  or  seems  so  -well  adapted  for  general  use,  as  that  above 
given..  This  rule  is   the  same  in  effect  as  Dp.  Halley's  rational 

formula. 


CUBE  ROOT.  87 

2.  Then  say,  by  the  Rule  of  Three,  As  th6  sum  of  the  givea 
number  and  double  the  assumed  cube,  is  to  the  sum  of  the 
assumed  cube  and  double  the  given  number,  so  is  the  root  of 
the  assumi'd  cube,  to  the  root  required,  nearly.  Or,  As  the 
fir^t  sum  is  to  the  difference  of  the  given  and  assumed  cube, 
so  IS  the  assumed  root  to  the  difference  of  the  roots  nearly. 

3.  Again,  by  using,  in  like  manner,  the  cube  of  the  root 
last  found  as  a  new  assumed  cube,  another  root  will  be  obtain- 
ed still  nearer.  And  so  on  as  far  as  we  please  ;  using  always 
the  cube  of  the  last  found  root,  for  the  assumed  cube. 

EXAMPLE. 

To  find  the  cube  root  of  21035-8. 

Here  we  soon  find  that  the  root  lies  between  20  and  30,  and 
then  between  27  and  28.     Taking  therefore  27,  its  cube  is 
19683,  which  is  the  assumed  cube.     Then 
19683  21035-8 

2  2 


39366 
21036-8 

42071-6 
19683 

60401-8  :  61764-6  : 
27 

:  :  27  :  27-6047. 

4322822 
1236092 

60401-8)  1667374-2  (27-6047  the  root  nearly. 

469338 

36626 

284 

42 


formula)  but  more  commodiously  expressed  ;  and  the  first  investiga- 
tion of  it  was  given  in  my  Tracts,  p.  49.  The  algebraic  form  of  it 
is  this  : 

As  p  +  2a  :  A  +  2p  : :  r :  R.    Or, 

As  p  4"  2-A-  :  p    c/J    A  :  :  r  ;  R   c/J  r  ,- 
where  p  is  the  given  number,  a  the  assumed  nearest  cube,  r  the  cube 
root  of  A,  and  r  the  root  of  p  sought* 

Again, 


88  ARITHMETIC. 

Again,  for  a  second  operation,  the  cube  of  this  root  is 
21035-318645155823,  and  the  process  by  the  latter  method 
will  be  thus  : 

21035-318646,  &c. 


42070-637290       21035-8 

21035-8  21035-318G45,  &c. 


As  63106-43729       :         diff.  481355  :  :  27-6047  : 

thediff.  -000210560 


conseq.  the  root  req.  is  27-604910560. 
Ex.  2.  To  extract  the  cube  root  of  -67. 
Ex.  3.  To  extract  the  cube  root  of -01. 

TO  EXTRACT  ANY  ROOT  WHATEVER. 


Let  p  be  the  given  power  or  number,  n  the  index  of  the 
power,  A  the  assumed  power,  r  its  root,  r  the  required  root 
of  p.     Then  say, 

As  the  sum  of  w  -f-  1  times  a  and  ?i  —  1  times  p^ 
is  to  the  sum  of  n  -{-  1  times  p  and  n  —  1  times  a  ; 
so  is  the  assumed  root  r,  to  the  required  root  R. 

Or,  as  half  the  said  sum  of  n  -j-  1  times  a,  and  n  —  1  times 
p,  is  to  the  difference  between  the  given  and  assumed  powers, 
so  is  the  assumed  root  r,  to  the  difference  between  the  true 
and  assumed  roots  ;  which  difference,  added  or  subtracted,  as 
the  case  requires,  gives  the  true  root  nearly. 

Thati8,n-|-1.  A+ n — 1.  p  :  n-fl.p. -f*n — 1.  a  ::  r  :  r. 

Or, n+  \.\A-\-n — ^^  1.  |p  :  p  c/2  a  :  ;  r  :  r  co  r. 
And  the  operation  may  be  repeated  as  often  as  we  please, 
by  using  always  the  last  found  root  for  the  assumed  root,  and 
its  nth  power  for  the  assumed  power  a. 


♦  This  is  a  very  g-eneral  approximating  rule,  of  which  that  for  the 
cube  root  is  a  particular  case,  and  is  the  best  adapted  for  practice, 
and  fnr  memory,  of  any  that  I  have  yet  seen.  It  was  first  discoveu-ed 
in  this  form  by  myself,  and  the  investigation  and  uge  of  it  were  given 
at  large  in  my  Tracts^  p.  45,  &€• 

EXAMPLE. 


GENERAL  ROOTS, 


89 


EXAMPLE. 
To  extract  the  5th  root  of  21035-8. 

Here  it  appears  that  the  5tli  root  is  between  7  3  and  7-4. 
Takinj?7-3,  its  6th  power  is  20730-71693.  Hence  we  have 
p  =  21036-8,  n  =  5  r  =  7-3  and  a  =  20730-71593  ;  then 


n  -f-  1 .  i  A-f-n  —  l.^Ptpco   A  ::  r  :  Rco   r,  that  is, 
3  X20730-71693  +2  X  21035-8   :  305  084  :  :  7-3  : 
3  2  7-3 


62192-14779              42071*6        915252 

42071-6                                       2135688 

(04263-74779                           ^227-1 132 (-02I3605=R,  co 

7-3= 

:r,  add. 

7-321360=R,true 

to  the  last  figure. 

OTHER  EXAMPLES. 

1. 

.  What  is  the  3d   root  of  2  ? 

Ans. 

1-25992L 

2, 

.  What  is  the  3d    root  of  3214  ? 

Ans. 

14-75758. 

3. 

What  is  the  4th  root  of  2  ? 

Ans. 

1-189207. 

4. 

What  is  the  4th  root  of  97-41  ?, 

Ans. 

3-1415999. 

5. 

What  is  the  5th  root  of  2  ? 

Ans. 

1-148699. 

6. 

What  is  the  6th  root  of  21035-8  ? 

Ans. 

5-254037. 

7. 

What  is  the  6th  root  of  2  ? 

Ans. 

1-122462. 

8. 

What  is  the  7th  root  of  21036-8  ? 

Ans. 

4-145392. 

9. 

What  is  the  7th  root  of  2  ? 

Ans. 

1  104089. 

10. 

What  is  the  8th  root  of  21035-8  ? 

Ans. 

3  470323. 

11. 

What  is  the  8th  root  of  2  ? 

Ans. 

1-090508. 

12. 

What  is  the  9th  root  of  21036-8  ? 

Ans. 

3-022239. 

13. 

What  is  the  9th  root  of  2  ? 

Ans. 

1-080069. 

The  following  is  a  Table  of  squares  and  cub^s,  as  also  the 
'square  roots   and  cube  roots,  of  all  numbers  from  1  to  1000, 
which  will  be  found  very  useful  on  many  occasions,  in  numeral 
calculations,  when  roots  or  powers  are  concerned. 


Vol.  I, 


13 


90      A  TABLE  OF  SQUARES,  CUBES,  AND  ROOTS. 


Number. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

1 

1 

1 

1  0000000 

i-oooooo 

2 

4 

8 

1-4142136 

1-259921 

3 

9 

27 

1-7320508 

•1-442250 

4 

16 

64 

2-0000000 

1-587401 

5 

25 

125 

2-2360680 

1-709976 

6 

36 

216 

2-4494897 

1-817121 

7 

49 

343 

2-6457513 

1-912933 

8 

64 

512 

2-8284271 

2*000000 

9 

81 

729 

3-0000000 

2*080084 

10 

loo 

1000 

3  1622777 

2-154435 

11 

121 

1331 

3-31G6248 

2223980 

12 

144 

1728 

3-4641016 

2-289428 

13 

169 

2197 

3-6055513 

2-351335 

14 

196 

2744 

3-7416574 

2-410142 

15 

225 

3375 

3-8729833 

2'466212 

16 

256 

4096 

4-0000000 

2-519842 

17 

289 

4913 

4-1231056 

2-571282 

18 

324 

5832 

4-2426407 

2-620741 

19 

361 

6859 

4-3588989 

2  668402 

20 

400 

8000 

4-4721360 

2-714418 

21 

441 

9.261 

4-5825757 

2-758923 

22 

484 

10648 

4-6904158 

2-802039 

23 

529 

12167 

4-7958315 

2-843867 

24 

576 

13824 

4-8989795 

2-884499 

25 

625 

15625 

5-0000000 

2-924018 

26 

676 

17576 

50990195 

2-962496 

27 

729 

19683 

5-1961524 

3-000000 

28 

784 

21952 

5-2915026 

3-036589 

29 

841 

24389 

5-3851648 

3072317 

30 

900 

27000 

5-4772256 

3-107232 

31 

961 

29791 

5-5677644 

3  141381 

32 

1024 

32768 

5-6568642 

3-174802 

33 

1089 

35937 

5-7445626 

3-207534 

34 

1156 

39304 

5-8309519 

3-239612 

35 

1225 

42875 

5-9160798 

3-271066 

36 

1296 

46656 

6-0000000 

3301927 

37 

1369 

50653 

6-0827625 

3-332222 

38 

1444 

54872 

6  1644140 

3-361975 

39 

1521 

59319 

6  2449980 

,  3-39121 1 

40 

1600 

64000 

6-3245553 

3-419952 

41 

1681 

68921 

6-4031242 

3  448217 

42 

1764 

74088 

64807407 

3-476027 

43 

1849 

79507 

6-5574385 

3-503398 

44 

1959 

85184 

6-6332496 

3  530348 

45 

2025 

91125 

6-7082039 

3556893 

46 

2116 

97336 

6-7823300 

3  583048 

47 

2209 

103823 

6-8556546 

3-608826 

48 

2304 

1 10592 

6*9282032 

3-634241 

49 

2401 

117649 

7-0000000 

3659306 

50 

2500 

125000 

7-0710678 

3-684031 

SQUARES,  CUB^S,  AND  ROOTS. 


91 


Number. 

Square. 

2601 

.  C«be. 

Square  Hoot. 

Cube  Koot. 

51 

132651 

7  1414284 

3  708430 

52 

2704 

140608 

7-2 111026 

3-732511 

53 

2809 

148877 

7  2801099 

3-756286 

54 

2916 

157464 

7-3484692 

3779763 

55 

3025 

166375 

74161985 

3-802953 

56 

3136 

175616 

7-4833148  > 

3-825862 

57 

3249 

185193 

7-5498344 

3  848501 

58 

3364 

195112 

7-6157731 

3  870877 

59 

3481 

205379 

7-6811457 

3-892996 

60 

3600 

216000 

7-7459667 

3-914867 

61 

3721 

226981 

7  8102497 

3-936497 

62 

3844 

238328 

7-8740079 

3-957892 

63 

3969 

250047 

7-9372539 

3*979057 

64 

•4096 

262144 

8  0000000 

4-000000 

65 

4225 

274625 

8-0622577 

4-020726 

66 

4356 

287496 

8-1240384 

4-041240 

67 

4489 

300763 

8'1853528 

4  061548 

68 

4624 

314432 

8-2462113 

4081656 

69 

4761 

328509 

8-3066239 

4  101566 

70 

4900 

343000 

8-3666003 

4-121285 

71 

5041 

357911 

8-4261498 

4  140818 

72 

5184 

373248 

8-4852814 

4-160168 

73 

5329 

389017 

8-5440037 

4 179339 

74 

5476 

405224 

8  6023253 

4-198336 

75 

5625 

421875 

8-6602540 

4-217183 

76 

5776 

438976 

5  7177979 

4-235824 

77 

5929 

456533 

8-7749644 

4-254321 

78 

6084 

474552 

8-8317609 

4272659 

79 

6241 

493039 

8  8881944 

4  290841 

80 

-6400 

5 1 2000 

8  9442719 

4-308870 

81 

6561 

531441 

9  0000000 

4326749 

82 

6724 

551368 

9  0553851 

4-344481 

83 

6889 

571787 

9-1104336 

4-362071 

&4 

7056 

592704 

9-1651514 

4-379519 

85 

7225 

614125 

9-2195445 

4-396830  . 

86 

7396 

"636056 

9-2736185 

4-414005 

87 

7569  • 

658503 

9  3273791 

4  431047 

88 

7744 

681472 

9-3808315 

4-447960 

89 

7921 

704969 

9-4339811 

4-464745 

90 

8100 

720000 

9  4868330 

4-481405 

91 

8281 

753571 

9o393920 

4-497942 

92 

8464 

778688 

9  5916630 

4514357 

93 

8649 

804357 

9-6436508 

4  530655 

94 

8836' 

830584 

9  6953597 

4  546836 

95 

9025 

857375 

9-7467943 

4-562903 

9*6 

9216 

*  884736 

9  7979590 

4578857 

97 

9409 

912673 

9-8488578 

4-594701 

98 

9604 

941192 

9-8994949 

4-610436 

99 

9801 

970299 

99498744 

4-626055 

100 

10000 

1000000 

10-0000000 

4  641589 

92 


ARITHMETIC. 


lui 

Square. 

Cube. 

Square  Root. 

Cube  Root 

10201 

1030:;oi 

10  0498756 

4  657010 

102 

10404 

1061208 

100995049 

4  672330  . 

103 

10609' 

1092727 

10-1488916 

4-687548 

104 

10816 

1124364 

10  1980390 

4-702669 

105 

1  ^9^5 

1157625 

10  2469508 

4717694 

106 

1  lf^6 

1191016 

10  2456301 

4-732624 

i07 

11449 

1225043 

10-3940804 

4  747459 

108 

11664 

1259712 

10-3923048 

-  4  762203 

109 

11881 

1295029 

10  4403065 

4-776856 

110 

12100 

1331000 

10-4880885 

4-791420 

111 

12321 

1367631 

10  5356538 

4-80589.6 

112 

12544 

1404928 

10-5830052 

4  820284 

113 

12769 

1442897 

I0'6301458 

4-834588 

114 

12996 

1481544 

10-6770783 

•  4-848808 

115 

13225 

1520875 

107238053 

4-862944 

116 

13456 

156<'896 

10-7703296 

4-876999 

117 

13689 

16U1613 

10-8166538 

4-890973 

118 

13924 

1643032 

108627805 

r4-904868 

119 

14161 

1685159 

109087121 

4-918685 

120 

14400 

1728000 

10  9544512 

4-932424 

121 

14641 

1771561 

11  0000000 

4946088 

122 

14884 

1815848 

11-0453610 

4  959675 

123 

15129 

1860867 

11-0905365 

4-973190 

124 

15376 

1906624 

111355287 

4986631 

125 

15625 

1953125 

11-1803399 

5*000000 

126 

15876 

2000376 

11  2ia49722 

5013298 

127 

16129 

2048383 

11-2694277 

5-026526 

128 

16384 

2097152 

11  3137085 

5  039684 

129 

16641 

2146689 

11  3578167 

5  052774 

150 

16900 

2197000 

11  4017543 

5-065797 

131 

17161 

2248091 

11-4455231 

5-078753 

132 

17424 

2299968 

11-4891253 

5091643 

133 

17689 

2352637 

11  5325626 

5  104469 

134 

17956 

2406104 

11-5758369 

5-117230 

*  135 

J  8225 

2460375 

ll'6i89500 

5  129928 

136 

18496 

2515456 

11  66190;58 

5-142563 

13r 

18769 

2571353 

11-7046999 

5-155137 

138 

19044 

2628072 

11-7473444 

5  167649 

139 

19321 

2685619 

11  7898261 

5-180101 

140 

19600 

2744000 

11-8321596 

5  192494 

141 

19881 

2803221 

11 '8743421 

5204828 

142 

20164 

2863288 

11>9163753 

5-217103 

143 

20449 

2924207 

11-9582607 

5229321 

144 

20736 

2985984 

12-0000000 

5241482 

145 

21025 

3048625 

120415946 

5  25358S 

146 

.  21316 

3112136 

12-0830460 

5  265637 

147 

2160-^ 

3176523 

12-1243557 

5-277632- 

148 

21904 

3241792 

12-1655251 

5-289572 

149 

22201  • 

330794a 

12-2065556 

5  301459 

150  1 

22500 

3375000 

12-2474487 

5-313293 

SQUARES,  CUBES,  AND  ROOTS. 


93 


Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

151 

22801 

3442951 

122882057 

5-325074 

152 

23104 

3511808 

123288280 

5-336803 

153 

23409 

3581577 

12  3693169 

5  348481 

154 

23716 

3652264 

12-4096736 

5  360108 

155 

24025 

3723875 

12-4498996 

5-371685 

156 

24336 

3796416 

12  4899960 

5  383213 

157 

24649 

3869893 

12  5299641 

5o94G90 

158 

24964 

3944312 

12-5698051 

5  4U6120 

159 

25281 

4019679 

126095202 

5-417501 

160 

25600 

4096000 

12-6491106 

5-428835 

161 

25921 

4173281 

12  6885775 

5-440122 

162 

26244 

4251528 

12-7279221 

5-451362 

163 

26569 

4330747 

12-7671453 

5-462556 

164 

26896 

4410944 

12-8062485 

5  475703 

165 

27225 

4492125 

12-8452326 

5-484806 

166 

27556 

4574296 

128840987 

5  495865 

167 

27889 

4657463 

12-9228480 

5-506879 

168 

28224 

4741632 

12-9614814 

5-517848 

169 

28561 

4826809 

13-0000000 

5  528775 

170* 

28900 

49 1 3000 

15-0384048 

5-539658 

171 

29241 

5000211 

130766968 

5-550499 

172 

29584 

5088448 

13-1148770 

5-561298 

173 

29929 

5177717 

13-1529464 

5-572054 

174 

30276 

5268024 

13-1909060 

5'582770 

175 

30625 

5359375 

13  2287566 

5-593445 

176 

30976 

545  1776 

13'2664992 

5 '604079 

177 

31329 

5545133 

•13  3041347 

5614673^ 

178 

31684 

5639752 

13  3416641 

5-625226 

179 

3204 1 

5735339  . 

13  3790882 

5-635741 

180 

32400 

5832000 

•  134164079 

5-646216 

181 

32761 

5929741 

13-4536240 

5-656652 

182 

33124 

6028568 

'   134907376 

5  667051 

183 

33480 

6128487 

13  5277493 

5-677411 

184 

33856 

6229504 

13-5646600 

5-687734 

185 

34225 

6331625 

13-6014705 

5-698019 

186 

34596 

6454856 

13-6381 8 '7 

5708267 

187 

34969 

6539203 

13  6747943 

57 18479 

138 

35344 

6644672 

13  7113092 

•5'728654 

189 

35721 

6751269 

13  7477271 

5-738794 

190 

36100 

6859000 

13-7840488 

5-748897 

191 

36481 

6967871 

13-8202750 

5-7589&5 

192 

36864 

7077888 

13-8564065 

5  768998 

193 

37249 

7189057 

15-8924440  . 

5  778996 

194 

37636 

7301384 

13-9283883 

5-788960 

195 

38025 

7414875 

13'-9642400 

.  5-798890 

196 

38416 

7529536 

14-0000000 

5-808786 

197 

38809 

7645373 

140356688 

5-818648 

198 

39204 

7762392 

14'0712473 

5-828476 

199 

39601 

7880599 

14-1067360 

5-838272 

200 

40000 

8000000 

14-1421356 

5  848035 

y4 

ARITHMETIC. 

Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

~  201 

40401 

8120601 

14M774469 

5-857765 

202 

40804 

8242408 

14-2126704 

5-867464 

203 

41209 

8365427 

14-2478068 

5-8771*30 

204. 

41616 

8489664 

14-2828569 

5-886765 

<   205 

42025 

8615125 

J4-3178211 

5-896368 

206 

42436 

8741816 

143527001 

5905941 

207 

42849 

8869743 

14-3874946 

5-915481 

208 

43264 

8998912 

14-4222051 

5-924991 

209 

43681 

9123329 

14-4568323 

5-934473 

210 

44  i  00 

9261000 

14-4913767 

5-9439  li 

211 

44521 

9393951 

14-5258390 

5-953341 

212 

44944 

9528128 

14-5602198 

5-962731 

213 

45369 

9663597 

14-5945195 

5-972091 

214 

45796 

9800344 

14-6287388 

5981426 

2J5 

46225 

9938375 

14-6628783 

5-909727 

216 

46656 

10077696 

14-6969385 

6-000000 

2ir 

47089 

10218313 

14-7309199 

6-009244 

218 

47524 

10360232 

14-7648231 

6-018463 

219 

47961 

10503459 

14-7986486 

6  027650 

220 

48400 

10648000 

14  8323970 

6036811 

221 

48841 

10793861 

14-8660687 

6-045943 

222 

49284 

10941048 

14-8996644 

6-055048 

223 

49729 

11089567 

14  9331845 

6-064126 

224 

50176 

11239424 

14-9666295 

6-073177 

225 

50625 

11390625 

15  0000000 

6  082201 

226 

51076 

11543176 

,   15  0332964 

6-091199 

227 

51529 

11697083 

15-0665192 

6-100170 

228 

51984 

11852352 

15-0996689 

6-109115 

229 

52441 

12008989 

15  1327460 

6-118032 

230 

52900 

12167000 

15-1657509 

6  126925 

231 

53361 

12326391 

15-1986842 

6-135792 

232 

53824 

12487168 

15-2315462 

6-144634 

233 

54289 

12649337 

15-2643375 

6-153449 

234 

54756 

12812904 

15-2970585 

6-162239 

235 

55225 

12977875 

15-3297097 

6-171005 

236 

55696 

13144256 

15-3622915 

6-179747 

237 

56169 

13312053 

15-3948043 

6-188463 

238 

56644 

13481272 

15-4272486 

6-197154 

239 

57121 

13651919 

15-4596248 

6  205821 

240 

57600 

13824000 

15-4919334 

6  214464 

241 

58081 

13997521 

15-5241747 

6-223083 

242 

58564 

14172488 

15-5563492 

6-231678 

243  . 

59049 

14348907 

155884573 

6-240251 

244 

59536 

14526784 

15-6204994 

6  248800 

245 

60025 

14706125 

15-^524758 

6  257324 

246 

60516 

14886936 

15-6843871 

6  265826 

247 

61009 

15069223 

15-7162336 

6-274304 

248i 

61504 

15252992 

15-7480157 

6-282760 

249 

62001 

15438249 

15-7797338 

6-291194 

250 

62500 

15625000 

15-8113883 

6299604 

SQUARES,  CUBES,  AND  ROOTS. 


Numb. 
~25T~ 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

63001 

15813251 

15-8429795 

.6-307992 

252 

63504 

16003008 

15*8745079  * 

6316359 

253 

64009 

16194277 

15-9059737 

6-324^04 

'2S4> 

64516 

16387064 

15-9373775 

^•333C'25 

255 

65025 

16581375 

15-9687194 

6-341355 

256 

65536 

16777216 

J  6 -.0000000 

6-349a02 

25/ 

66049 

16974593 

160312195 

6^357859 

258 

66564 

17173512 

16-0623784 

6  366095 

259 

67081 

17373979 

16'0934769 

6-374310 

260 

67600 

17576000 

16-1245155 

6-382504 

261 

68121 

17779581 

16-1554944 

6-390676 

262 

68644 

17984728 

16-1864141 

6-398827 

263 

69169 

18191447 

16-2172747 

6406958 

264 

69696 

18399744 

16-2480768 

6-415068 

265 

70225 

18609625 

16-2788206 

6-423157 

266 

70756 

18821096 

16-3095064 

6431226 

267 

71289 

19034163 

16-3401346 

6-439275 

268 

71824 

19248832 

16-3707055 

6447305 

269 

72361 

19465109 

164012195 

6  455314 

270 

72900 

19683000 

1.64316767 

6463304 

271 

73441 

19902511 

16  4620776 

6  471274 

272 

73984 

20123648 

16-4924225 

6-479224 

273 

74529 

20346417 

1(5  5227116 

6-487153 

274 

75076 

20570824 

15-5529454 

6-495064 

275 

75625 

20796875 

155831240 

6  502956 

276 

76176 

21024576 

!  6-6 132477 

6-510829 

277 

76729 

21253933 

16-6433170 

6-518684 

278 

77284 

21484952 

16-6733320 

6-526519 

279 

77841 

21717639 

167032931 

6-534335 

280 

78400 

21952000 

16-7332005 

6  542132 

281 

78961 

22188041 

16-7630546 

6-549911 

282 

79524 

22425768 

16  7928556 

6-557672 

283 

80089 

22665187 

16-8226038 

6-565415 

284 

80656 

22906304 

16-8522995 

6-573139 

285 

81225 

2314912^ 

168819430 

6-580844 

286 

81796 

23393656 

16-9115345 

6-588531 

287 

82369 

23639903 

16  9410743 

6-596202 

288 

82944 

23887872 

16-9705627 

6-603854 

289 

83521 

24137569 

17  0000000 

6  611488 

290 

84100 

24389000 

170293864 

6.619106 

291 

84681 

24642171 

170587221 

,  6  626705 

292 

85264 

24897088 

17  0880075 

6-634287 

293 

85849 

25153757 

171172428 

6-641851 

294 

86436 

25412184 

.  17-1464282 

.  6-649399 

295 

87025 

25672375 

17-1755640 

6656930 

296 

87616 

25934336 

17-2046505 

.  6-664443 

297 

88209 

26198073 

17-2336879 

6-671940 

298 

88804 

26463592 

17-2626765 

6  679419 

299 

89401 

26730899 

17-2916165 

6-686882 

300 

90000 

^7000000 

17-3205081 

6-694328 

96 


ARITHMETIC. 


Numb. 

301 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

90601 

27270901 

17-34935  16 

6  701758 

302 

91204 

27543608 

17'3781472 

6-709172 

303 

91809 

27818127 

17  4068952 

6  716569 

304 

92416 

28094464 

17-4355958 

6-723950 

305 

93025 

28372625 

17-4642492 

6-7313  6 

306 

93636 

28652616 

17-4928557 

6-738665 

307 

94249 

28934443 

17-5214155 

6-745997 

308 

94864 

29218112 

17-5499288 

6  753^13 

309 

95481 

29503629 

17*5783958 

6-7606,4 

310 

96100 

29791000 

17-6068169 

6767899 

311 

96721 

30080231 

17-6351921 

6775168 

312 

97344 

30371328 

17-6635217 

6  782422 

313 

97969 

30664297 

17-6918060 

6  789661 

314 

98596 

30959144 

17  7200451 

6  796884 

■  315 

99225 

3:255875 

17-7482393 

6804091 

316 

99856 

31554496 

17  7763888 

6-811284 

317 

100489 

31855013 

17-8044938 

6818461 

318 

101124 

32157432 

178325545 

6-825624 

319 

101761 

32451759 

17-8605711 

6-832771 

320 

102400 

32758000 

17-8885438 

6-839903 

321 

103041 

330r6161 

.  17-9104729 

6-847021 

322 

103684 

33386248 

17-9443584 

6-854124 

323 

104329 

33698267 

17-9722008 

6*861211 

324 

104976 

34012224 

18-0000000 

6*868284 

325 

105625 

34328125- 

18-0277564 

6-875343 

326 

106276 

34645976 

18  0554701 

6-882388 

327 

106929 

34965^83 

18  0831413 

6-889419 

328 

107584 

35287552 

18  1107703 

6-896435 

329 

108241 

35611289 

18-1383571 

6  903436 

330 

108900 

35937000 

18-1659021 

6-910423 

331 

109561 

36264^91 

18-1934054 

6-917396 

332 

110224 

36594568 

18-2208672 

6-924355 

333 

110889 

369260S7 

18-2482876 

6-931300 

334. 

111556 

37259704 

"18-2756669 

6  938232 

335 

112225 

376953r5 

18-3030052 

6  945149 

336 

112896 

•37933056 

18-3303028 

6  952053 

337 

113569 

38272753 

18-3575598 

6-958943 

338 

114244 

58614472 

18-4847763 

6.9658j9 

339 

114921 

38958219 

18  4119526 

6  972682 

340 

115600 

39304000 

184390889 

6.979532 

341 

116281 

39651821 

18^4661853 

6.986369 

342 

116964 

40001688 

18-4932420 

6-993  591 

343 

117649 

40353607 

18-5202592 

7-000000 

344 

118336 

407X)7584 

18*5472370 

7«006796 

345 

119025 

41063625 

18'5741756 

7-013579 

346 

119716 

41421736 

186010752 

7-020349 

347 

120409 

41781923 

18  6279360 

7  027106 

,  348 

121104 

42144192  i 

18'654758l 

r-e33850 

349 

121801 

42508549 

18*6815417 

7-040581 

350 

122500 

43875000 

18*7082869 

7047208 

Si^UARES,  CUBES,  AND  ROOTS. 


97 


Numb. 

j  Square. 

Cube. 

Square  Root. 

Cube  Root 

351 

123201 

43243551 

187349940 

7-054003 

352 

123904 

43614208 

18-7616630 

7-060696 

353 

124609 

43986977 

18-7882942 

7-067376 

354 

125316 

44361864 

18-8148877 

7-074043 

355 

126025 

44738875 

18-8414437 

7  080698 

356 

126736 

45118016 

18-8679623 

7087341 

357 

127449 

45499293 

18-8944436 

7093970 

358 

128164 

45882712 

18  9208879 

7-100588 

359 

128881 

46268279 

18-9472953 

7-107193 

360 

129600 

46656000 

18-9736660 

7  113786 

361 

130321 

47045881 

19'00Q0000 

7*120367 

362 

131044 

47437928 

19-0262976 

7*126935 

363 

131769 

47832147 

19-0525589 

7  133492 

364 

132496 

48228544 

19-0787840 

7*140037 

365 

133225 

48627125 

19-1049732 

7-146569 

366 

133956 

49027896 

19-I3!i265 

7*153090 

367 

134689 

49430863 

19-1572441 

7-159599 

368 

135424 

49836032 

19-1833261 

7-166095 

369 

136161 

50243409 

19*2093727 

7  172580   ' 

370 

136900 

50653000 

19-2353841 

7-179054 

371 

137641 

51064811 

19-2613603 

7-185516 

372 

138384 

51478848 

19-2873015. 

7-191966 

373 

139129 

51895117 

19*3132079 

7-198405 

374* 

139876 

52313624 

19*3390796 

7  204832 

375 

140625 

52734375 

19-3649167 

7-211247 

376 

141376 

53157376 

19-3907194 

7  217652 

377 

142129 

53582633 

19*4164878 

7*224045 

378 

142884 

54010152 

194422221 

7*230427 

379 

143641 

54439939 

194679223 

7-236797 

380 

144400 

54872000 

19-4935887 

7-243156 

381 

145161 

55306341 

19-5192213 

7-249504 

382 

145924 

55742968 

19-5448203 

7  255841 

283 

146689 

56181887 

19-5703858 

7*262167 

384 

147456 

56623104 

19-5959179 

7-268482 

385 

148225 

57066625 

196214169 

7-274786 

386 

148996 

57512456 

19*6468827 

7-281079 

387 

149769 

57960603 

19-6723156 

7^87362 

388 

150544 

58411072 

19  6977156 

7293633 

389 

151321 

58863869 

19*7230829 

7-299S93 

390 

152100 

59319000 

19*7484177 

7-306143 

391 

152881 

59776471 

19-7737199 

7312383 

^92 

153664 

60236288 

19-7989899 

7-318611 

393 

154449 

60698457 

19-8242276 

7-3248i!9 

394 

155236 

61162984 

19  8494332 

7-331037 

395 

156025 

61629875 

19  8746069 

7-337234 

396 

156816 

62099136 

19-8997487 

7  343420 

397 

157609 

62570773 

19  9248588 

7-349596 

398 

158404 

63044792 

19-9499373 

7  355762 

399 

159201 

63521199 

199749844 

7-361917 

400 

160000 

64000000 

20-.0000000 

7-368063 

Vol.  1 


H 


^ 

ARITHMETIC. 

Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

401 

1 6080 1 

64481201 

20  0249844 

7-374198 

402 

J  6 1 604 

64964808 

200499377 

7-380322 

403 

162409 

65450827 

20  0748599 

7-386437 

404 

163216 

65939264 

20-0997512 

7-392542 

405 

164025 

66430125 

20-1246118 

7-398636 

406 

164836 

66923416 

20  1494417 

7-404720 

407 

165649 

67419143 

20-1742410 

7-410794 

408 

166464 

67911312 

20-1990099 

7-416859   ' 

409 

167281 

68417929 

20-2237484 

7-422914 

410 

, 168100 

68921000 

20-2484567 

7-428958 

411 

168921 

69426531 

20-2731349 

7  434993 

,  412 

169744 

69934528 

20-2977831 

7-441018 

413 

170569 

70444997 

20-3224014 

7-447033 

414 

171396 

70957944 

20-3469899 

u     7-453039 

415 

172225 

71473375 

203715488 

7-459036 

416 

173056 

71991296 

20-3960781 

7-465022 

417 

173889 

72511713 

.   20-4205779 

7-470999 

418 

174724 

73034632 

204450483 

7-476966 

419 

175561 

73560059 

20-4694895 

7-482924 

420 

176400 

74088000 

20-4939015 

7488872 

421 

177241 

74618461 

20-5182845 

7-494810 

422 

178084 

75151448 

20-5426386 

7-500740 

423 

178929 

75686967 

20-5669638 

7*506660 

424 

179776 

76225024 

20*5912603 

7512571 

425 

180625 

76765625 

20  6155281 

7-518473 

426 

181476 

77308776 

20-6397674 

7-524365 

427 

182329 

77854483 

20-663978^3 

7-530248 

428 

1^5184 

78402752 

20-6881609 

7-536121 

429 

184041 

78953589 

20-7123152 

7-541986 

430 

184900 

79507000 

20-7364114 

7-547841 

431 

185761 

80062991 

20-7605395 

7-553688 

'  432 

186624 

80621568 

20-7846097 

7-559525 

-  433 

187489 

81182737 

208086520 

7-565353 

434 

■188356 

81746504 

20-8326667 

7-571173 

435 

189225 

82312875 

20-8566536 

7-576984 

-  436 

190096 

82881856 

20-8806130 

7-582786 

437 

190969 

83453453 

20  9045450 

7*588579 

438 

191844 

84027672 

30  9284495 

7-594363 

439 

192721 

84604519 

209^523268 

7-600138 

440 

193600 

85 1  84t)00 

20-9761770 

7-605905   • 

441 

194481 

S5766121 

210000000 

7-6 11662 

442 

195364 

86350888 

21-0237960 

7-617411 

443 

196249 

86938307 

21  0475652 

7-623151 

444 

197136 

87528384 

210713075 

^  7628883 

445 

198025 

88121125 

21-0950231 

7-634606 

446 

198916 

88716536 

2ril87l21 

7-640321 

447 

199809 

89314623 

21-1423745 

7-646027 

448 

200704 

89915392 

21-1660105 

r-651725 

.  449 

201601 

90518849 

21-1896201 

7-657414 

450 

202500 

91125000 

21  2132034 

7-663094  * 

SQUARES,  CUBES,  ANB  ROOTS. 


?5? 


;  Numb. 

Squire. 

Cube. 

Square  Root. 

Cube  Root. 

451 

203401 

91733851 

21-2367606 

7-668766 

452 

204304 

92345408 

21-2602916 

7-6744:50 

453 

205209 

92959677 

21-2837967 

7.680085 

454 

206116 

93576664, 

21  3072758 

7»685732 

455 

207025 

94196375 

21-3307290 

7-691371 

456 

207936 

94818816 

21-3541565 

7-697002 

457 

208849 

95443993 

21-3775583 

7-702624 

458 

209764 

96071912 

21-4009346 

7-708238 

459 

210681 

96702579 

21-4242853 

7  713844 

460 

2U600 

97336000 

21-4476106 

7-719442 

461 

212521 

97972181 

21-4709106 

7-725032 

462 

213444 

98611128 

21  4941853 

7  730614 

463 

214369 

99252847 

21-5174348 

7-736187 

454 

215296 

99897344 

21-5406592 

7-741753 

465 

216225 

100544625 

21-5638587 

7-747310 

466 

217156 

101194696 

21-5870331 

7-752860 

457 

218089 

101847563 

21-6101828 

7-758402 

468 

219024 

102503232 

21-6333077 

7-763936 

469 

219961 

103161709 

21-6564078 

7-769462 

470 

220900 

103823000 

21-6794834 

7'774980 

471 

221841 

104487111 

21-7025344 

7-780490 

472 

222784 

105154048 

21-7255610 

7-785992 

473 

223729 

105823817 

21-7485632 

7-791487 

474 

224676 

106496424 

21-7715411 

7-796974 

475 

3256'^5 

107171875 

21-7944947 

7-802453 

476 

226576 

107850176 

21-8174242 

7-807925 

477 

227529 

108531333 

21-8403297 

7-813389 

478 

228484 

109215352 

21-8632111 

7.818845 

479 

229441 

109902239 

21-8860686 

7-824294 

480 

230400 

110592000 

21-9089023 

7-829735 

48i 

231361 

Ill284ii41 

21-9317122 

7-835168 

432 

1^32324 

111980168 

21-9544984 

7.840594 

483 

233289  " 

■112678587 

21-9772610 

7-846013 

484 

234256 

113379904 

22-0600000 

7-851424 

485 

235225 

114084125 

22  0227155 

7-856828 

486 

•236196 

114791256 

22-0454077 

7-862224 

487 

237169 

115501303 

22-0680765 

7-867613 

488 

238144 

116214272. 

22*0907220 

7  872994 

489 

239121 

116930169 

22- !  133444 

7.878368 

490 

240100 

117849000 

22- 1359436 

7. 883734 

491 

241081 

118370771 

221585198 

7.889094 

492 

242064 

119095488 

221810730 

7-894446 

493 

243049 

119823157 

22-2036033 

7-899791 

494 

244036 

120553784 

22  2261408 

7  905129 

495 

24502  5 

121287375 

22-2485955 

7  910460 

.  496 

246016 

122023936 

22-2710575 

7  915784 

497 

24r009 

122763473 

22-2934968 

7-921100 

498 

248004 

123505992 

22-3159136 

7-926408 

499 

249001  ' 

124251499 

22  338S079 

7  931710 

500 

250000 

125000000 

22*3606798 

7-937005  ' 

100 

AftlTHMETlC. 

Numb. 

Square. 

Cube. 

Square  Koot. 

22  3830293 

Cube  Root. 

501 

251001 

125751501 

7-942293 

502 

252004 

126506008 

22-4053565 

7-947573 

503 

253009 

127263527 

224276615 

7-952847 

504 

254016 

128024064 

22-4499443 

7-958114 

505 

255025 

128787625 

22-4722051 

r-963374 

506 

256036 

129554216 

22.4944438 

7-968627 

507 

257049 

130323843 

22-5166605 

7-973873 

508 

258064 

131096512 

22-5388553 

7-979 112 

• 

509 

259081 

131872229 

22-5610283 

7-984344 

510 

260100 

132651000 

22-5831796 

7-989569 

511 

261121 

133432831 

22-6053091 

7-994788 

512 

262144 

134217728 

22-6274170 

8-000000 

513 

263169 

135005697 

22  6495033 

8005205 

514 

264196 

135796744 

22-6715681 

8-010403 

515 

265225 

136590875 

22  6936114 

8015595  . 

516 

266256 

137388096 

22  7156334 

8  020779 

5ir 

267289 

138188413 

22-7376340 

8-025957 

518 

268324 

138991832 

22-7596134 

8-031129 

519 

269361 

139798359 

22-78  J  5715 

8-036293 

520 

270400 

140608000 

228035085 

8  041451 

521 

271441 

141420761 

22-8254244 

8046603 

522 

272484 

142236648 

22  8473193 

8051748 

523 

273529 

143055667 

22-8691933 

8-056886 

524 

274576 

143877824 

22  8910463 

8*062018 

525 

275625 

144703125 

22-9128785 

8-067143 

526 

276676 

145531576 

22-9346899 

8-072262 

527 

277729 

146363183 

22  9564806 

8-077374 

528 

278784 

147197952 

22-9782506 

8082480 

529 

279841 

148035889 

23-0000000 

8-087579 

530 

280900 

148877000 

23-0217289 

8-092672 

531 

281961 

149721291 

23  0434372 

8-097758 

532 

283024 

150568768 

23-0651252 

8-102838 

533 

284089 

151419437 

23-0867928 

8-107912 

534 

285156 

152273304 

23-1084400 

8-112980 

535 

286225 

153130375 

23-1300670 

8-118041 

536 

287296 

153990656 

23-1516738 

8-123096 

537 

288369 

154854153 

23-1732605 

8-128144 

538 

289441 

155720872 

23-1948270 

8-133186 

539 

290521 

156590819 

23  2163735 

8*138223 

540 

291600 

157464000 

23  2379001 

8*143253 

541 

.  292681 

158340421 

23-2594067 

8*148276 

542 

293764 

159220088 

23-2808935 

8' 153293 

543 

294849 

160103007 

23-3023604 

8-158304 

544 

295936 

160989184 

23-3238076 

8  163309 

545 

297025 

161878625 

233452351 

8-168308 

546 

298116 

162771336 

23-5666429 

8-173302 

547 

299209 

163667323 

23  3880311 

8-178289 

548 

300304 

164566592 

234093998 

8-183269 

549 

301401 

165469149 

23-4307490 

8-188244 

550 

302500 

1C6375000 

23*4520788 

8-193212 

SQUARES,  CUBES,  ANt>  ROOTS. 


to? 


Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

"  551  ~ 

303601 

167284151 

23-4733892 

8-198175 

552 

304704 

168196608 

23-4946802 

8-203131 

553 

305809 

169112377 

23-5159520 

8  208082 

554 

306916 

170031464 

23-5372046 

8-213027 

555 

308025 

170953875 

23-5584380 

8-217965 

556 

309136 

171879616 

23-5796522 

8-222898 

557 

310249 

172808693 

23  6008474 

8-227825 

558 

311364 

173741112 

23-6220236 

8-232746 

659 

312481 

174676879 

236431808 

8-237661 

560 

343600 

175616000 

23-6643191 

8-242570 

561 

314721 

176558481 

23-6854386 

8-247474 

562 

315844 

177504328 

23-7065392 

8-252371 

563 

316969 

178453547 

23-7276210 

8-257263 

564 

318096 

179406144 

23-7486842 

8-262149 

^65 

319225 

180362125 

23-7697286 

8-267029 

566 

320356 

181321496 

23-7907545 

8-271903 

567 

321489 

182284263 

23-8117618 

8-276772 

568 

322624 

183250432 

23-8327506 

8281635 

569 

323761 

184220009 

23-8537209 

8-286493 

570 

324900 

185193000 

23-8746728 

8-291344 

571 

326041 

186169411 

23-8956063 

8-296190 

572 

327184 

187149248 

23-9165215 

8-301030 

573 

328329 

188132517 

23-9374184 

8-305865 

574 

329476 

189119224 

23-9582971 

8'310694 

575 

3306"25 

190109375 

23-9791576 

8*315517 

576 

331776 

191102976 

24*0000000 

8*320335 

'  577 

332929 

192100033 

24-0208243 

8*325147 

578 

334084 

193100552 

24-0416306 

8-329954 

579 

335241 

194104539 

24-0624188 

8-334755 

580 

336400 

195112000 

24-0831892 

8-339551 

581 

337561 

196122941 

24-1039416 

8-344341 

582 

338724 

197137368 

24  1246762 

8-349125 

583 

339889 

198155287 

241453929 

8-353904 

584 

341056 

199176704 

24-1660919 

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585 

342225 

200201625 

24-1867732 

8-363446 

586 

343396 

201230056 

24-2074369 

8-368209 

587 

344569 

202262003 

24-2280829 

8-372966 

538 

345744 

203297472 

24-2487113 

8-377718 

589 

346921 

204336469 

24-2693222 

.8382465 

590 

348100 

205379000 

24-2899156 

8-387206 

591' 

349281 

206125071 

24-3104916 

8-391942 

592 

350464 

207474688 

24-3310501 

8  396673 

593 

351649 

208527857 

243515913 

8-401398 

594 

352836 

209584584 

24-3721152 

8  406118 

595 

354025 

210644875 

24  3926218 

8-410832 

596 

355216 

211708736 

24-4131112 

8-415541  , 

597 

356409 

212776173 

24-4335834 

8  420245 

598 

357604 

213847192 

24-4540385 

8-424944 

599 

358801 

214921799 

24-4744765 

8-429638 

600 

360000  * 

216000000 

24-4948974 

8  434327 

■'M'- 


AKITHMETIG. 


Numb. 

Square. 

Cube. 

Square  Root 

Cube  Root." 

601 

361201 

217081801 

24-5153013  • 

8^439009 

602 

362404 

218167208 

24^5356883 

8-443687 

603 

363609 

219256227 

24-5560583 

8-448360  . 

604 

364816 

220348864 

24'5764115 

8-4*3027  ' 

605 

366023 

221445125 

24-5967478 

8-457689 

606 

367236 

222"545016 

24' 6 170673 

8-462347 

607 

368449 

223643543 

24.637S700 

8-466999 

608 

369664 

224756712 

24-6576560 

8*471647 

609 

37088i 

225866529 

24-6779254 

8*476289 

610 

372100 

226981000 

24-6981781 

81^480926 

611 

373321 

228099131 

24-7184142 

8*485557 

612 

374544 

229220928 

24-7386338 

8-490184 

613 

375769 

230346397 

24-7588368 

8*494806 

614 

376996 

231475544 

24-7790234 

8-499423 

615 

378225 

232.608375 

24-7991935 

8-504034 

616 

379456 

253744896 

24-8193473 

8'608641 

617 

380689 

234885113 

24-8394847 

8-513243 

618 

381924 

236029032 

24-8596058 

8-517840 

619 

383161 

237176659 

24-8797106 

8-5224v32 

620 

384400 

238328000 

24-8997992 

8-527018 

621 

385641 

239483061 

24-9198716 

8-531600 

622 

386884 

240641848 

24-9399278 

8-536177 

623 

388129 

241804367 

24-9599679 

8-540749 

,  624 

389376 

242970624 

24*9799920 

8-545317 

625 

390625 

244140625 

25-0000000 

8-549879 

626 

391876 

245314376 

25-0199920 

8-554437 

627 

393129 

246491883 

'  25-0399681 

8-558990 

628 

394384  . 

247673152 

25-0599282 

8-563537 

629 

395641 

248858189 

25-0798724 

8-568080 

630 

396900 

250047000 

25  0998008 

8-572618 

631 

398161 

251239591 

25-1197134 

8-577152 

632 

399424 

252435968 

25-1396102 

8-581680 

633 

400689 

253636137 

251594913 

8-586204 

634 

401956 

254840104 

25-1793566 

8-590723 

635 

403225 

256047875 

25-1992063 

8-595238 

636 

404496 

257259456 

25-2190404 

8-599747 

637 

405769 

258474853 

25-2388589 

8  604252 

638 

407044 

259694072 

25-2586619 

8-608753 

639 

408321 

260917119 

25-2784493 

8-613248 

640 

409600 

262144000 

25  2982213 

8617738 

641 

410881 

263374721" 

25-3179778 

8622224 

642 

412164 

264609288 

25-3377189 

8  626706 

643 

413449 

265847707 

25-3574447 

8-631183 

644 

414736 

267089984 

25  3771551 

.  8  635655 

645 

416025 

268336125 

25-3968502 

8  640122 

646 

417316 

269586136 

25-4165301 

8  644585 

647 

418609 

270840023 

25  4361947 

8-649043 

648 

419904 

272097792 

25-4558441 

8  653497 

649 

421201 

273359449 

25-4754784 

8-657946 

650 

422500 

274625000 

25*4950076 

8-662301 

SQUARES,  CUBES,  AND  ROOTS. 


io3 


Numb. 
651 

^Square. 

Cube. 

Square  Hoot. 

Cube  Roet. 

433801 

275894451 

25  5147016 

8-666831 

652 

425104 

277167808 

25-5342907 

8-671266 

653 

426409 

278445077 

S5-5538647 

8675697 

654 

427716 

279726264 

25-5734237 

8*680123 

655 

429025 

281011375 

25-5929678 

8  684545 

656 

430336 

282300416 

25-6124969 

8-688963 

657 

431649 

283593393 

25-6320112 

8-693376 

658 

432964 

284890312 

'^5  65\5lOr 

8-697784 

659 

434281 

286191179 

25-^709953 

8-702188 

660 

435600 

287496000 

25-6904652 

8  706587 

661 

436921 

288804781 

25-7099203 

8  710982 

662 

438244 

290117528 

25-7203607 

8  715373 

663 

439569 

291434247 

25-7487864 

8*719759 

664 

440896 

292754944 

25-7681975 

8-724141 

665 

442225 

294079625 

25-7875939 

8-728518 

666 

443556 

295408296 

25-8069758 

8-732891 

667 

444889 

296740963 

25-8263431 

8-737260 

668 

446224 

298077632 

25-8456960 

8-741624 

669 

447561 

299418309 

•25-8650343 

8-745984 

670 

448900 

360763000 

25-8843582 

8-750340 

671 

450241  ■ 

302111711 

25-9036677 

8-754691 

672 

451584 

303464448 

25-9229628 

8-759038 

673 

452929 

304821217 

25-9422435 

8-763380 

674 

454276 

306182024 

25' 96 1 5 100 

8-767719 

675 

455625 

307546875 

25-9807621 

8-772053 

676 

456976 

308915776 

26  0000COO 

8-776382 

677 

458329 

310288733 

26-0192237 

8-780708 

678 

459684 

311665752 

26-0384331 

8-785029 

679 

461041 

313046839 

26-0576284 

8-789346 

680 

462400 

314432000 

260768096 

8-793659 

681 

463761 

315821241 

26-0959767 

8-797967 

682 

465124 

317214568 

26-1151297 

8-802272 

683 

466489 

318611987 

261342687 

8-806572 

684 

467856 

320013504 

26  1535937 

8-810868 

685 

469225 

321419125 

26-1725047 

8-815159 

686 

470596 

322828856 

26.1916017 

8819447 

687 

471969 

324242703 

26-2106848 

8-823730 

688 

473344 

325660672 

26-2297541 

8-828009 

689 

474721 

327082769 

26-2488095 

8-832285. 

690 

476100 

328509000 

26-2678511 

8-836556 

691 

47748 r 

329939371 

26  2868789 

8-840822 

692 

478864 

331373888 

26-3058929 

8-845085 

693 

480249 

332812557 

26-3248932 

8  849344 

694 

481636 

334255384 

26-3438797 

8-853598 

695 

483025 

335702375 

26-3628527 

8-857849 

696 

484416 

337153536 

263818119   . 

8-862095 

697 

485809 

338608873 

26-4007576 

8-866337 

698 

487204 

340068392 

26*4196896 

8-870575 

699 

488601 

341532099 

26-4386081 

8-874809 

700 

490000 

343000000 

26-4575131 

8-879040 

104 

ARITHMETIC. 

Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Hoot. 

701 

491401 

344472101 

26-4764046 

8-88  3266 

702 

492804 

345948008 

26*4952826 

8-887488 

703 

494209 

347428927 

265141472 

8-891706 

704 

495616 

348913664 

26-5329983 

8'895920 

705 

497025 

350402625 

26-5518361 

8-900130 

706 

498436 

351895816 

26-5706605 

8-904336 

707 

499849 

353393243 

265894716 

8-908538 

708 

501264 

354894912 

26-6082694 

8-912736 

709 

502681 

356400829 

26-6270539 

8-916931 

710 

504100 

357911000 

26*6458252 

8-921121 

711 

505521 

359425431 

26-6645833 

8925307 

712 

506944 

360944128 

26*6833281 

,8'929490 

713 

508369 

362467097 

26-7020598 

8-933668 

714 

509796 

363994344 

26-7207784 

3'937843 

715 

511225 

365525875 

26-7394839 

8-942014 

716 

512656 

367061696 

26-7581763 

8-946180 

717 

514089 

368601813 

26-7768557 

8-950343 

718 

515524 

370146232 

26-7955220 

8-954502 

719 

516961 

371694959 

26-8141754 

8*958658 

720 

518400 

373248000 

26-8328157 

8-962809 

721 

519841 

374805361 

26-8514432 

8-966957 

722 

521284 

376367048 

26-8700577 

8-971100 

723 

522729 

377933067 

26-8886593 

8-975240 

724 

524176 

379503424 

26-9072481 

8*979376 

725 

525625 

381078125 

26  9258240 

8-983508 

726 

527076 

382657176 

26*9443872 

8'987637 

727 

528529 

384240583 

26-9629375 

8991762 

728 

529984 

385828352 

26-9814751 

8*995883 

729 

531441 

387420489 

27-0000000 

9  000000 

730 

532900 

389017000 

27-0185122 

9*004113 

731 

534361 

390617891 

27-0370117 

9-008222 

732 

535824 

392223168 

27-0554985 

9*012328 

733 

537289 

393832837 

27-0739727 

9-016430 

734 

538756 

396446904 

27-0924344 

9020529 

735 

540225 

397065375 

27-1108834 

9-024623 

736 

541696 

398688256 

27-1293199 

9*028714 

737 

543169 

400315553 

27-1477439 

9  032802 

738 

544644 

401917272 

27-1661554 

9-036885 

739 

546121 

403583419 

27-1845544 

9-040965 

740 

547600 

405224000 

27-2029410 

9  045041 

741 

549081 

406869021 

27-2213152 

9  049114 

742 

550564 

'408518488 

27-2396769 

9-053183 

743 

552049 

410172407 

27-2580263 

9-057248 

744 

553536 

411830784 

27  2763634 

9061309 

745 

555025 

413493625 

27-2946881 

9  065367 

746 

556516 

415160936 

27-3130006 

9-069422 

747 

558009 

416832723 

27-3313007 

9-073472 

748 

559504 

418508992 

27-3495887 

9-077519 

749 

561001 

420189749 

27-3678644 

9-081563 

750 

662500 

421875000 

27-3861279 

9-085603 

SQUARES,  CUBES,  AND  ROOTS. 


105 


Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

751 

564001 

423564751 

27-4043792 

9*089639 

752 

565504 

425259008 

27-4226184 

9093572 

753 

567009 

426957777 

27-4408455 

9097701 

754 

568516 

428661064 

274  590604 

9  101726 

755 

570025 

4303688? 5 

27-4773633 

9  105748 

756 

571536 

432081216 

27  4954542 

9-109766 

757 

573049 

433798093 

27  5136330 

9  113781 

758 

574564 

435519512 

27  5317998 

9.117793 

759 

576081 

437245479 

27-5499546 

9-121801 

760 

577600 

438976000 

27-5680975  . 

9-125805 

761 

579121 

440711081 

275862284 

9-129806 

762 

580644 

442450728 

27  6043475 

9- 133303 

763 

582169 

444194947 

27  6224546 

9  137797- 

764 

583696 

445943744 

27-6405499 

9  141788 

765 

585225 

447697125 

27-6586334 

9-145774 

766 

586756 

449455096 

276767050 

9-149757 

767  , 

588289 

451217663 

27-6947648 

9-153737 

768 

589824 

452984832 

27-7128129 

9  157713 

769 

591361 

454756609 

27-7308492 

9  161686 

770 

592900 

456533000 

27-7488739  - 

9  165656 

771 

594441 

458314011 

27-7668868 

9-169622 

772 

595984 

460099648 

27-7848880 

9  17S585 

773 

597529 

461889917 

27-8028775 

9-177544 

774 

599076 

463684824 

27*8208555 

9  181500 

775 

600625 

465484375 

27-8388218 

9-185452 

776 

602176 

467288576 

27-8567766 

9  189401 

777 

603729 

469097433 

27-8747J97 

9-193347 

778 

605284 

470910952 

27-8926514 

9  197289 

779 

606841 

472729139 

27-9105715 

9  201228 

780 

y608400 

474552000 

27-9284801 

9-205164 

781 

609961 

476379541 

27-9463772 

9  209096 

782 

611524 

478211768 

27-9642629 

9  213025 

783 

613089 

480048687 

27-9821372 

9-216950 

784 

614656 

481890304 

28-0000000 

9220872 

785 

616225 

483736025 

280178515 

9-224791 

786 

617796 

485587656 

28-0356915 

9228706 

787 

619369 

487443403 

28-0535203 

9  232618 

788 

620944 

485303872 

28-0713377 

9237527 

789 

622521 

491169069 

28-0891438 

'9*240433 

790 

624100 

493039000 

28-1069386 

9  244335 

791 

625681 

494913671 

28-1247222 

9-248234 

792 

627264 

496793088 

28  1424946 

9-252130 

793 

628849 

498677257 

28-1602557 

9  256022 

794 

630436 

500566184 

28  1780056 

9  259911 

795 

632025 

502459875 

28-1957444 

9-263797 

796 

633616 

504358336 

28-2134720 

9-267679 

797 

635209 

506261573 

'28-2311884 

9-271559 

798 

636804 

508169592 

28  2488938 

9-275435 

799 

638401 

510082399 

28  2665881 

9-279308 

800 

640000 

512000000 

28-2842712 

9-283177 

Vol.  I 


15 


306 


ARITHMETIC. 


Numl). 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

■  801 

641601 

513922401 

28-3019434 

9-287044 

803' 

643204 

515849608 

28-3196045 

9-290907 

803 

644809 

517781627 

28-3372546 

9-294767 

804 

646416 

519718464 

28-3548938 

9-298623 

805 

64802.5 

521660125 

28-3725219 

9-302477 

806 

649636 

523606616 

28-3901391 

9-S06327 

807 

651249 

525557943 

28-4077454 

9310175 

808 

652864 

527514112 

28-4253408 

9-314019 

809 

654481 

529475129 

28-4429253 

9317859 

810 

656100 

531441000 

28-4604989 

9-321697 

811 

657721 

533411731 

28-4780617 

9-325532 

812 

659344 

535387328 

28-4956137 

9-329363 

813 

660969 

537366797 

28-5131549 

9-333191 

814 

662596 

539353144 

28-5306852 

9-337016 

815 

664225 

541343375 

28-5482048 

9-340838 

816 

665856 

543338496 

28-5657137 

9344657 

817 

667489 

545338513 

28-5832119 

9348473 

818 

669124 

547343432 

28*6006993  . 

9  362285 

819 

670761 

549353259 

28-6181760 

9-356095 

820 

672400 

551368000 

28  6356421 

9-359901 

821 

674041 

553387661 

28-6530976 

9-363704 

822 

675684 

555412248 

28  6705424 

9-367505 

823 

677329 

557441767 

28-6879766 

9*371302 

824 

678976 

559476224 

28*7054002 

9-375096 

825 

680625 

561515625 

28*7228132 

9-378887 

826 

682276 

563559976 

28-7402157 

9  382675 

827 

683929 

565609283 

287576077 

^-386460 

828 

685584 

567663552 

28-7749891 

9-390241 

829 

687241 

569722789 

28-7923601 

9-394020 

830 

688900 

571787000 

28-8097206 

9  397796 

831 

690561 

573856191 

28-8270706 

9-401569 

832 

692224 

575930368 

28  8444102 

9-405338   . 

833 

693889 

578009537 

28-8617394 

9-409105 

834 

695556 

580093704 

28-8790582 

9  412869 

835 

697225 

582182875 

28-8963666 

9-416630 

836 

698896 

584277056 

28-9136646 

9  420387 

837 

700569 

586376253 

28-9309523 

9-424141 

838 

702244 

588480472 

28  9482297 

9*427893 

839 

703921 

590589719 

28  9654967 

9-431642 

840 

705600 

592704000 

28  9827535 

9-435388 

841 

707-281 

594823321 

29-0000000 

9-439130 

«42 

708964 

596947688 

29  0172363 

9  442870 

843 

710649 

599077107 

29-0344623 

9-446607 

844 

712336 

601211584 

29-0516781 

9-450341 

845 

714025 

603351125 

29-0688837 

9*454071 

846 

715716 

605495736 

29  0860791 

9  457799 

847 

717409 

607645423 

29-1032644 

9-461524 

848 

719104 

609800192 

29-1204396 

9465247 

849 

720801 

611960049 

29-1576046 

9  468966 

I  850 

722500 

614125000 

29*1547505 

9-472682 

SQUARES,  CUBES,  AND  ROOTS. 


Wl 


Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

851 

724201 

616295051 

29-1719043 

9-476395 

852 

725904 

618470208 

29  1890390 

9-480106 

853 

727609 

620650477 

29  2061637 

9  483813 

854 

729316 

622835864 

29  2232784 

9-487518 

855 

731025 

625026375 

292403830 

9  491219 

856 

732736 

627222016 

29-2574777 

9494918 

857 

734449 

629422793 

29-2745623 

9  498614 

858 

736164 

631628712 

29  2916370 

9  502307 

859 

737881 

633839779 

293087018 

9-505998 

860 

739600 

636056000 

29  3257566 

9  509685 

861 

741321 

638277381 

29-3428015 

9  513369 

862 

743044 

640503928 

29  3398365 

9  517051 

863 

744769 

642735647 

29-3768616 

9  520730 

864 

746496 

644972544 

29  3938769 

9  524406 

865 

748225 

647214625 

29  4108823 

9  528079 

866 

749956 

649461896 

29  4278779 

9-531749 

867 

751689 

651714363 

29-4448637 

9  535417 

868 

753424 

653972032 

29-4618397 

9-539081 

869 

755161 

656234909 

29-4788059 

9  542748 

870 

756900 

658503000 

29-4957624 

9-546402 

871 

758641 

660776311 

29  5127091 

9-550058 

872 

760384 

663054848 

29-5296461 

9  553712 

873 

762129 

665338617 

29-5465734 

9-557363 

874 

763876 

667627624 

29  5634910 

9-561010 

875 

765625 

669921875 

29  5803989 

9  564655 

876 

767376 

672221376 

29  5972972 

9  568297 

877 

769129 

674526133 

29-6141858 

9-571937 

878 

770884 

676836152 

29  6310643 

9-575574 

879 

772641 

679151439 

29  6479325 

9  579208 

880 

774400 

681472000 

29-6647939 

9-582839 

881 

776161 

683797841 

29  6816442 

9-586468 

882 

777924 

686128968 

296984848 

9-590093 

883 

779689 

688466387 

29-7153159 

9  593716 

884 

-781456 

690807104 

29-7321375 

9  597337 

885 

783225 

693154125 

29-7489496 

9G00954 

886 

784996 

695506456 

29-7657521 

9-604569 

887 

786769 

698764103 

29-7825452 

9-608181 

888 

788544 

700227072 

29-7993289 

9-611791   ^ 

889 

790321 

702595369 

29  8161030 

9.615397 

890 

792100 

704969000 

29-8328678 

9*619001 

891 

793881 

707347971 

29-8496231 

9-622603 

892 

795664 

709732288 

29  8663690 

9  626201 

893 

797449 

712121957 

29  8831056 

9  629797 

894 

799236 

714516984 

298998328 

9  633390 

895 

801025 

716917375 

29  9165506. 

9-636981 

896 

802816 

719323136 

29  9332591 

9  640569 

897 

804609 

721734273 

29-9499583 

9  644154 

898 

806404 

724150792 

29  9666481 

9-647736 

899 

808201 

726572699 

29  9833287 

9  651316 

900 

810000 

729000000 

30  0000000 

9  654893 

lOS 

ARITHMETIC. 

Numb. 

Square 

Cube. 

Square  Root. 

Cube  Root. 

901 

811801 

731432701 

300166620 

9  658468 

902 

813604 

733870808 

300333148 

9  662040 

903 

815409 

736314327 

30-0499584 

9;  665  609 

904 

817216 

738763264 

300665928 

9-669176 

905 

819025 

741217625 

30-0832179 

9  672740 

906 

820836 

743677416 

S-^  0998339 

9-676301* 

907 

822649 

746142643 

3u  1164407 

9-679860 

908 

824464 

748613312 

30' 1330383 

9-6834!  6 

909 

826281 

751089429 

30- 1496269 

9  686970 

910 

828100 

753571000 

30  1632063 

9-690521 

911 

829921 

756058031 

30  1827765 

9-694069 

912 

831744 

7585o0528 

30  1993377 

9-697615 

913 

833569 

761048497 

302158899 

9  701158 

914 

835396 

763551944 

30-2:>24329 

9  704698 

9)5 

837225 

76G060875 

30*2489669 

9  708236 

916 

839056 

768575296 

30  2654919 

9-711772 

917 

840889 

771095213 

30  2820079 

9-7153Q5 

918 

842724 

773620632 

30-2985148 

9.718835 

919 

844561 

776151559 

30-3150128 

9-722363 

920 

846400 

778688000 

30  3315018 

9-725888 

92t 

8'48241 

781229961 

30  3479818 

9-729410 

922 

850084 

783777448 

30  3644529 

9-732930 

923 

851929 

786330467 

303809151 

9-736448 

•924 

P53776 

7888-89024 

30  3973683  . 

9-739963 

925 

855625 

791453125 

30  4138127 

9-743475 

926 

857476 

794022776 

30'4302481 

9-746985 

927 

859329 

796597983 

30  4466747 

9  750493 

928 

861184 

799178752 

304630924 

9  753998 

929 

& 63041 

801765089 

304795013 

9-757500 

930 

864900 

804357000 

30-4959014 

9-761000 

931 

866761 

806954491 

30  5122926 

9-764497 

932 

868624 

809557568 

30  5286750 

9-767992 

933 

870489 

812166237 

SO  5450487 

9-771484 

934 

872356 

814780504 

30-5614136 

9-774974 

935 

874225 

817400375 

30  5777697 

9-778461 

936 

876096 

820025856 

30*5941171 

9-782946 

937 

877969 

822656953 

SO  6104557 

9-785428 

938 

879844 

825293672 

30'6267857 

9-788908 

939 

881721 

827936019 

30  6431069 

9-792386 

940 

883600 

830584000 

30  6594194 

9795861 

941 

885481 

833237621 

30/6757233 

9-799333 

942 

887364 

835896888 

30-6920185 

9' 802803 

943 

889249 

838561807 

30  7083051 

9-806271 

944 

891136 

841232384 

30-7245830 

9-809736 

945 

893025 

843908625 

30-7408523 

98i3198 

946 

894916 

846590536 

30  7571130 

9  816659 

947 

896809 

849278123 

30-7733651 

9-820117 

948 

898704 

851971392 

30-7896086 

9'823572 

949 

900601 

854670349 

30-8058436 

9-827025 

950 

902500 

857375000 

30  8220700 

9-830475 

SQUARES,  CUBES,  AND  ROOTS. 


109 


Numb. 

Square. 

Cube. 

Square  Root. 

Cube  Root. 

951 

9044bT 

"  86008~535r 

30-8382879 

9-833923 

952 

906304 

862801408 

30-8544972 

9-837369 

953 

908209 

865523177 

30-8706981 

9-840812 

954 

910116 

868250664 

30*8868904 

9-844253 

955 

912025 

870985yS75 

30-9030743 

9-847692 

956 

913936 

873722816 

309192497 

9-851128 

957 

915849 

876467493 

30-9354166 

9-85456h 

958 

917764 

87921791-2 

30-9515751 

9-857992 

959 

919681 

881974079 

30-9677251 

9-861421 

960 

921600 

884736000 

30-9838668 

9*861848 

961 

923521 

887503681 

31-0000000 

9-868272 

96-2 

925444 

890277128 

31-0161248 

9-871694 

963 

927369 

8.93056347 

31-0322413 

9  875113 

964 

929296 

895841344 

31-0483494 

9-878530 

965 

931225 

89;i632125 

31-0644491 

9-881945 

966 

933156 

901428696 

31-0805405 

9-885357 

967 

935089 

9.04231063 

31-0966236 

9-888767 

968 

937024 

907039232 

31-1126984 

9  892174 

969 

938961 

909853209 

3 M 287648 

9-895580 

970 

,940900 

912673000 

311448230 

9-898983 

971 

942841 

915498611 

3M608729 

9-902383 

972 

944784 

918330048 

31  1769145 

9-905781 

973 

946729 

921167317 

31  1929479 

9-909177 

974 

948676 

924010424 

31-2089731 

9-912571 

975 

950625 

926859375 

31-2249900 

9-915962 

976 

952576 

929714176 

312409987 

9-919351 

977 

.954529 

932574833 

31-2569992 

9-922738 

978 

956484 

935441352 

.31-2729915 

9926122 

979 

958441 

938313739 

31  2889757 

9-929504 

980 

960400 

941192001  - 

31-3049517 

9-932883 

981 

962361 

944076141  . 

31-3209195 

9-936261 

982 

964324 

946966168 

31-3368792 

9-939636 

983 

966289 

949862087 

31-3528308 

9  943009 

984 

968256 

952763904 

31-3687743 

9-946379 

985 

970225 

95567)625 

31-3847097 

9-949747 

986 

972196 

958585256 

31-4006369 

9'953113 

987 

974169 

961504803 

31*4165561 

9-956477 

988 

976144 

964430272 

31*4324673 

9-959839 

989 

978121 

967361669 

31*4483704 

9-963198 

990 

980100 

970299000 

31*4642654 

9  966554 

991 

982081 

973242271 

31*4801525 

9-969909 

992  , 

984064 

976191488 

31-4960315 

9-973262 

993 

986049 

979146657 

31  5119025 

9-976612 

994 

988036 

982107784 

31-5277655 

9  979959 

995 

990025 

985074875 

31-5436206 

9  983304 

996 

992016 

988047936 

31-5594677 

9  986648 

997 

994009 

991026973 

31.5753068 

9  989990 

998 

996004 

994011992 

31-5911380 

9-993328 

999 

998001 

997002999 

31-6069613 

9996665 

no  ARiTHMETiC. 


OF  RATIOS,  PROPORTIONS,  AND  PROGRESSIONS. 

Numbers  are  compared  to  each  other  in  two  different 
tvays  :  the  one  comparison  considers  the  difference  of  the  two 
numbers,  and  is  named  Arithmetical  Relation;  and  the  dif- 
ference sometimes  the  Arithmetical  Ratio  :  the  other  consi- 
ders their  quotient,  which  is  called  Geometrical  Relation  ; 
and  the  quotient  is  the  Geometrical  Ratio.  So,  of  these  two 
numbers  6  and  3,  the  difference,  or  arithmetical  ratio,  is  6—3 
or  3,  but  the  geometrical  ratio  is  |  or  2. 

There  must  be  two  numbers  to  form  a  comparison  :  the 
number  which  is  compared,  being  placed  first,  is  called  the 
Antecedent ;  and  that  to  which  it  is  compared,  the  Conse- 
quent. So,  in  the  two  numbers  above,  6  is  the  antecedent, 
and  3  the  consequent. 

If  two  or  more  couplets  of  numbers  have  equal  ratios,  or 
equal  differences,  the  equahty  is  named  Proportion,  and  the 
terms  of  the  ratios  Proportionals.  So,  the  two  couplets,  4,  2 
and  8,  6,  are  arithmetical  proportionals,  because  4  —  2  =  JJ 
—  6  =  2;  and  the  two  couplets  4,  2  and  6,  3,  are  geometri- 
cal proportionals,  because  |.  |  =  2,  the  same  ratio. 

To  denote  numbers  as  being  geometrically  proportional,  a 
colon  is  set  between  the  terms  of  each  couplet,  to  denote  their 
ratio  ;  and  a  double  colon,  or  else  a  mark  of  equality  between 
the  couplets  or  ratios.  So,  the  four  proportionals,  4,  2,  6,  3 
are  set  thus,  4  :  2  :  :  6  ;  3,  which  means  that  4  is  to  2  as  6 
is  to  3  ;  or  thus,  4  :  2  =  6  :  3,  or  thus,  ±  z=z  &^  both 
which  mean,  that  the  ratio  of  4  to  2,  is  equal  to  the  ratio 
of  6  to  3. 

Proportion  is  distinguished  into  Continued  and  Disconti- 
nued. When  the  difference  or  ratio  of  the  consequent  of 
one  couplet,  and  the  antecedent  of  the  next  couplet,  is  not  the 
same  as  the  common  difference  or  ratio  of  the  couplets,  the 
proportion  is  discontinued.  So,  4,  2,  8,  6  are  in  discontinued 
arithmetical  proportion,  because  4  —  2  =  8  —  6  =  2^  where- 
as 8  —  2  =  6:  and  4,  2,  6,  3  are  in  discontinued  geometrical 
proportion,  because  f  =  |  =  2,  but  |  =  3,  which  is  not 
the  same. 

But  when  the  difference  or  ratio  of  every  two  succeeding 
terms  is  the  same  quantit}',  the  proportioa  is  said  to  be  Conti- 
nued, and  the  numbers  themselves  make  a  series  of  Continued 

Proportionals, 


ARITHMETICAL  PROPORTION.  ill 

Proportionals,  or  a  progression.  So  2,  4,  6,  8  form  an  arith- 
metical progression,  because  4—2  =  6—4  =  8—6  =  2,  all 
the  same  common  difference  ;  and  2,  4,  8,  16  a  geometrical 
progression,  because  f  =  |  V  =  2'  »*^  *^^  same  ratio. 

When  the  follov/ing  terms  of  a  progression  increase,  or 
exceed  each  other  it  is  called  an  Ascending  Progression,  or 
Series  ;  but  when  the  terms  decrease,  it  is  a  descending  one. 
So,  0,  f ,  2,  3,  4,  &c.  is  an  ascending  arithmetical  progression, 
but  9,  7,  6,  3,  1 ,  &c.  is  a  descending  arithmetical  progression. 
Also  1,  2,  4,  8, 16,  &c.  is  an  ascending  geometrical  progrebbion, 
and  16,  8,  4,  2,  1,  kc.  is  a  descending  geometrical  progression. 


ARITHMETICAL  PROPORTION  and  PROGRESSION. 

In  Arithmetical  Progression,  the  numbers  or  terms  have  all 
the  same  common  difference.  Also,  the  first  and  last  terms 
of  a  Progression,  are  called  the  Extremes  ;  and  the  other 
terms,  lying  between  them,  the  Means.  The  roost  useful 
part  of  arithmetical  propartions,  is  contained  in  the  follow- 
ing theorems  : 

Theorem  1.  When  four  quantities  are  in  arithmetical 
proportion,  the  sum  of  the  two  extremes  is  equal  to  the  sum 
of  the  two  means.  Thus,  of  the  four  2,  4,  6,  8,  here  2  + 
8  =  4  +  6  =  10. 

Theorem  2.  In  any  continued  arithmetical  progression, 
the  sum  of  the  two  extremes  is  equal  to  the  sum  of  any  two 
means  that  are  equally  distant  from  them,  or  equal  to  double 
the  middle  term  when  there  is  an  uneven  number  of  terms. 

Thus,  in  the  terms  1,  3,  5,  it  is  1  -f  5  =  3  -f-  3  =  6. 

And  in  the  series  2,  4,  6,  8,  10,  12,  14,  it  is  2  -f  14  «=  4 
-f  12  =  6  4-  10  =  8  + 8  =  16. 

Theorem  3.  The  difference  between  the  extreme  terms 
of  an  arithmetical  progression  is  equal  to  the  common  dif- 
ference of  the  series  multiplied  by  one  less  than  the  number 
of  the  terms.  So,  of  the  ten  terms,  2,  4,  6,  8,  10,  12,  14, 
16,  18,  20,  the  common  difference  is  2,  and  one  less  than 
the  number  of  terms  9  ;  then  the  difference  of  the  extremes 
is  20—2  ~  18,  and  2X9=18  also. 

Consequently, 


112  ARITHMETIC. 

Consequently,  the  greatest  term  is  equal  to  the  least  term 
added  to  the  product  of  the  conunon  difference  multiplied  by 
1  less  than  the  aumt>er  of  terms. 

Theorem  4.  The  sum  of  all  the  terms,  of  any  arithme- 
tical progression,  is  equal  to  the  sum  of  the  two  extremes  mul- 
tiplied by  the  number  of  terms,  and  divided  by  2  ;  or  the  sum 
of  the  two  extremes  multiplied  by  the  number  of  the  terms, 
gires  double  the  sum  of  all  the  terms  in  the  series. 

This  is  made  evident  by  setting  the  terms  of  the  series  in 
am  inverted  order,  under  the  same  series  in  a  direct  order,  and 
adding  the  corresponding  terms  together  in  that  order.  Thus^ 
in  the  series  1,  3,  6,  7,  9,  11,  13,  15. 
ditto  inverted  15,  13,  11,  9,  7,  5,  3,  1  ! 
.  the  sums  are  16  -f  16  -f  16  -f  16  -f  16  -f  16  -f  16  -f  16,  ' 
which  must  be  double  the  •  sum  of  the  single  series,  and  is 
equal  to  the  sum  of  the  extremes  repeated  as  often  as  are  the 
number  of  the  terms. 

From  these  theorems  may  readily  be  found  any  one  of 
these  five  parts  ;  the  two  extremes,  the  number  of  terms,  the 
common  difference,  and  the  sum  of  all  the  terms,  when  any 
three  of  them  are  given  ;  as  in  the  following  problems  : 

FEODLEM  I. 

Given  the   extremet,  and  the   Number  of   Tenns  ;  to  find  the 
Sum  of  all  the  Terms. 

Add  the  extremes  together,  multiply  the  sum  by  the  num- 
ber of  terms,  and  divide  by  2. 

EXAMPLES. 

1.    The  extremes    being  3  and    19,    and  the   number   of 
terms  9  ;  required  the  sum  of  the  terms  ? 
19 
3 


19-f3  22 

Or, X9  =  -^X9=llX9=i99. 

2  2 

the  same  answer. 


S.  It  is  required  to  find  the  number  of  all  the  strokes  a 
common  clock  strikes  in  one  whole  revolution  of  the  index, 
•r  in  12  hours  ?  Ans.  76. 

El. 


22 

9 

2) 

198 

Ans 

.99 

ARITHMETICAL  PROGRESSION.  113 

Ex.  3.  How  many  strokes  do  the  clocks  of  Venice  strike 
in  the  compass  of  the  day,  which  go  continually  on  from  1 
to  24  o'clock  ?  Ans.  300. 

4.  What  debt  can  be  discharged  in  a  year,  by  weekly  pay- 
ments in  arithmetical  progression,  the  first  payment  being  1», 
and  the  last  or  52d  payment  6/  3s  1  Ans.  135/  4«. 

PROBLEM  II. 

Given  the  Extremes,   and  the  Number  of  Terms ;  to  find  th$ 
Common  Difference. 

Subtract  the  less  extreme  from  the  greater,  and  divide 
the  remainder  by  1  less  than  the  number  of  terms,  for  the 
common  difference. 

EXAMPLES. 

1.  The  extremes  being  3  and  19,  and  the  number  of  terms 
9  ;  required  the  common  difEerence  ? 
19 
3  19—3       16 

8)  16  '    9—1       8 

Ans.    2 


2.  If  the  extremes  be  10  and  70,  and  the  number  of  terms 
21  ;  what  is  the  common  difference,  and  the  sum  of  the 
series  ?  Ans.  the  com.  diff.  is  3,  and  the  sum  is  840. 

3.  A  certain  debt  can  be  discharged  in  one  year,  by  weekly 
payments  in  arithmetical  progression,  the  first  payment  being 
Is,  and  the  last  bl  3s  ;  what  is  the  common  difference  of  the 
terms  ?  Aifs.  2, 

PROBLEM  m. 

Given  one  of  the  Extrtmes^  the  Common  Difference,  and  the 
JS'umher  of  Terms  :  to  find  the  other  Extreme,  and  the  sum 
of  the  Series. 

Multiply  the  common  difference  by  1  less  than  the  num« 
ber  of  terms,  and  the  product  will  be  the  difference  of  the 
extremes  :  Therefore  add  the  product  to  the  less  extreme,  to 
give  the  greater  ;  or  subtract  it  from  the  greater,  to  give  the 
less  extreme. 

Vol.  I.  16 

EXAMPI4ES, 


lU  ARITHMETIC.  . 

EXAMPLES. 

1 .  Given  the  least  term  3,  the  common  diflference  2,  of*  ans; 
arithmt^tkal  series  of  9  terms  ;  to  find  the  greatest  term,  and 
the  sum  of  the  series. 

2 

8 

16 


19  the  greatest  term> 
3  the  least  \ 

22  sum 
9  number  of  terms* 

*  2)  198 

99  the  sum  of  the  series. 

.  2.  If  the  greatest  term  be  70,  the  common  diflference  3y 
and  the  number  of  terms  21,  what  is  the  least  term,  and  the 
turn  of  the  series  ? 

Ans.  The  least  term  is  10,  and  the  sum  is  840; 

8.  A  debt  can  be  discharged  in  a  year,  by  paying  1  shilling 

the  first  week,  3  shillings  the  second,  and  so  on,  always  2 

■hillings  more  every  week  ;  what  is  the  debt,  and  what 'will 

^e  last  payment  be  ? 

Ans.  The  last  payment  will  be  6/  3s,  and  the  debt  is  135/  45. 
PROBLEM  IV. 
To  find  an  Arithmetiml  Mean  Proportional  between  Two  Given 
Terms. 

Adb  the  two  given  extremes  or  terms  together,  and  take 
half  their  sum  for  the  arithmetical  mean  required. 

EXAMPLE. 

To  find  an  arithmetical  mean  between  the  two  numbers  4 
and  14. 

Here 
14 
4 

2)  18 

Ana.    9  the  mean  required. 

PROBLEM 


ARITHMETICAL  PROGRESSION,  lU 

FHOBLEM  V. 

To  find  two  Arithmetical  Means  between  Two  Given  Extremes. 

Subtract  the  less  extreme  from  the  greater,  and  divide 
the  difference  by  3,  so  vfAl  the  quotient  be  the  common  dif* 
lerence  ;  which  being  continually  added  to  the  less  extreme^ 
•r  taken  from  the  greater,  gives  the  means. 

EXAMPU5. 

To  find  two  arithmetical  means  between  2  and  8. 
Here  8 


3  )  6         Then  2  -f-  2  =  4  the  one  mean, 
and  4  4-  2  =  6  the  other  mean. 


"^om.  dif.      2 


PROBLEM  VI. 


To  find  any  Numher  of  Arithmetical  Means  between  Two  Given^ 
Terms  or  Extremes. 

Subtract  the  less  extreme  from  th6  greater,  and  divide 
the  difference  by  1  more  than  the  number  of  means  required 
to  be  found,  which  will  give  the  common  difference  j  then 
this  being  added  continually  to  the  least  term,  or  subtracted 
from  the  greatest,  will  give  the  terms  required. 

EXAMPLE. 

To  find  five  arithmetical  means  between  2  and  14. 
Here  14 
2 

6)  12     Then  by  adding  this  com.  dif.  continually, 
- — ^        the  means  are  found  4,  6,  8,  10,  12. 
com.  dif.  ^ 


See  more  of  ArithmeticaJ  progression  in  the  Algebra. 

GEOMETRICAL 


H^  ARITHMETIC. 


GEOMETRICAL  PROPORTION  AND  PROGRESSION, 

In  Geometrical  Progression  the  numbers  or  terms  have 
all  the  same  multiplier  or  divisor.  The  most  useful  part  of 
Geometrical  Proportion  is  eoetaiaed  in  the  following  theo- 


Theorem  1.  When  four  quantities  are  in  geometrical  pro- 
portion, the  product  of  the  two  extremes  is  equal  to  the  pro- 
duct of  the  two  means. 

Thus,  in  the  four  2,  4,  3,  6,  it  is  2  X  6  =  3  X  4  =  12. 

And  hence,  if  the  product  of  the  two  means  be  divided  by 
©ne  of  the  extremes,  the  quotient  will  give  the  other  extreme. 
So,  of  the  a^bove  numbers,  the  product  of  the  means  12-7-2 
=  6  the  one  extreme,  and  12-^-^  =  2  the  other  extreme  ; 
and  this  is  the  foundation  and  reason  of  the  practice  in  the 
Rule  of  Three.   , 


Theorem  2.  In  any  continued  geometrical  progression,  the 
product  of  the  two  extremes  is  equal  to  the  product  of  any 
two  means  that  are  equally  distant  from  them,  or  equal  to  the 
square  of  the  middle  term  when  there  is  an  uneven  number 

01  terms. 

Thus,  in  the  terms  2,  4,  8,  it  is  2  X  8  =  4  X  4  =  16. 

And  in  the  series  2,  4,  8, 16,  32,  64,  128, 

it  is  2  X  128  =  4  X  64  =  8  X  32  =  16  X  16  =  256. 

THEORESf  3.  The  quotient  of  the  extreme  terms  of  a 
geometrical  progression,  is  equal  to  the  common  ratio  of  the 
series  raised  to  the  power  denoted  by  1  less  than  the  number 
of  the  terms.  Consequently  the  greatest  term  is  equal  to 
the  least  term  multiplied  by  the  said  quotient. 

So,  of  the  ten  terms,  2,  4,  8,  16,*32,  64,  128,  256,  612, 
1024,  the  common  ratio  is  2,  and  one  less  than  tbe  nuaiber 
of  terms  is  9  ;  then  the  quotient  of  the  extremes  is    1024-r- 

2  =  612,  and  2°  =  612  also. 

Theorem 


GEOMETRICAL  PROGRESSION.  117^ 

Theorem  4.  The  sum  of  all  the  terms,  of  any  geometri- 
cal progression,  is  found  by  adding  the  greatest  term  to  the 
difierence  of  the  extremes  divided  by  1  less  than  the  ratio. 

So,  the  sum  of  2,  4,  8,  16,  32,  64,  128,  256,  512,  1024, 
1024—2 

(whose  ratio  is  2),  is  1024  -\ =  10244-1022  =  2046. 

2—1 

^  The  focegoing,  and  several  other  properties  of  geometrical 
proportion,  are  demonstrated  more  at  large  in  the  Algebraic 
part  of  this  work.  A  few  examples  may  here  be  added  of  the 
theorems,  just  delivered,  with  some  problems  concerning  mean 
proportionals. 

EXAMPLES. 

1.  The  least  often  terms,  in  geometrical  progression,  being 
1,  and  the  ratio  2  ;  what  is  the  greatest  term,  and  the  sum  of 
all  the  terms  ? 

Ans.  The  greatest  term  is  512,  and  the  sum  1023. 

2.  What  debt  may  be  discharged  in  a  year  or  12  months,  by 
paying  11  the  first  month,  2/  the  second,  4/  the  third,  and  so  on, 
each  succeeding  payment  being  doublethe  last  ;  and  what  will 
the  last  payment  be  ? 

Ans.  The  debt  4095?,  and  the  last  payment  2048/. 


PROBLEM  L 

To  find  One  Geometrical  Mean  Proportional  between  any  Two 
JVumbers. 

Multiply  the  two  numbers  together,  and  extract  the  square 
root  of  the  product,  which  will  give  the  mean  proportional 
sought. 

EXAMPLE. 

To  find  a  geometrical  mean  between  the  two  numbers  3 
and  12. 

12 
3 


36  (6  the  mean. 

3G 


PROBLEM 


118  ARITHMETIC, 


PROBLEM  H. 


To  find  Two  Geometrical  Mean  Proportionals  between  any  Tagy 
Numbers. 

Divide  ^he  greater  number  by  the  less,  and  extract  the 
cube  root  of  the  quotient,  which  will  give  the  common  ratio 
of  the  terms.  Then  multiply  the  least  given  term  by  the  ratio 
for  the  first  mean,  and  this  mean  again  by  the  ratio  for  the  se- 
cond mean  :  or,  divide  the  greater  of  the  two  given  terms  by 
the  ratio  for  the  greater  mean,  and  divide  this  again  by  the 
ratio  for  the  less  mean. 


EXAMPLE. 

To  find  two  geometrical  means  between  3  and  24. 

Here  3)  24  (8  ;  its  cube  root  2  is  the  ratio. 

Then  3  X  2  =  6,  and  6  X  2  =  12,  the  two  means. 

Or  24  -r  2  =  12,  and  12  -f-  2  =  6,  the  same. 

That  is,  the  two  means  between  3  and  24,  are  6  and  12. 


PROBLEM  IIL 

To  find  any  Number  of  Geometrical  Means  between  Two  Nurh" 
bers. 

Divide  the  greater  number  by  the  less,  and  extract  such 
root  of  the  quotient  whose  index  is  1  more  than  the  number  of 
means  required  ;  that  is,  the  2d  root  for  one  mean,  the  3d  root 
for,  two  means,  the  4th  root  for  three  means,  and  so  on  ;  and 
that  root  will  be  the  common  ratio  of  all  the  terms.  Then, 
with  the  ratio,  multiply  continually  from  the  first  term,  or  di- 
vide continually  from  the  last  or  greatest  term. 


EXAMPLE 

To  find  four  geometrical  means  between  3  and  96. 

Here  3)  96  (  32  ;  the  5th  root  of  which  is  2,  the  ratio. 
Then  3  X  2  =  6,  &  6 X  2  =  12,  &  12 X2  =  24,  &  24 X2  =  48- 
Or  96-^2  =;  48,  &  48-f-2  =  24,  &  24—2  ==  12,  &  12-i-2  =  6. 

That  is,  6,  12, 24,  48,  are  fhe  four  means  between  3  and  96. 

OF 


MUSICAL  PROPORTION.  Ud 

OF  MUSICAL  PROPORTION. 


TvERE  is  also  a  third  kind  of  proportion,  called  Musical, 
which  being  but  of  little  or  no  common  use,  a  very  short  ac- 
count of  it  may  here  suffice. 

Musical  Proportion  is  when,  of  three  numbers,  the  first 
has  the  same  proportion  to  the  third,  as  the  difference  between 
the  first  and  second,  has  to  the  difference  between  the  second 
and  third. 

As  in  these  three,  6,  0,  12  ; 

where  6  :  12  :   :  8  —  6  :   12  —  8, 
that  is  6  :  12  :  :  2  :  4. 

When  four  numbers  are  in  musical  proportion  ;  then  the 
first  has  the  same  ratio  to  the  fourth,  as  the  difference  be- 
tween the  first  and  second  has  to  the  difference  between  the 
third  and  fourth. 

As  in  these,  6,  8,  12,  18  ; 
where  6:18::  8-6  :   18  —  12^. 
that  is  C  :  18  :  :  2  :  6. 

When  numbers  are  in  musical  progression,  their  recipro- 
cals are  in  arithmetical  progression  ;  and  the  converse,  that 
is,  when  numbers  are  in  arithmetical  progression,  their  reci- 
procals are  in  musical  progression. 

So  in  these  musicals  6,  8,  12,  their  reciprocals  },  i,  j\, 
are  in  arithmetical  progression  ;  for  ^  -f-  yL  =  jl.  =  i  j 
and  I  4-  i  =  I  =  i  ;  that  is,  the  sum  of  the  extremes  is 
equal  to  double  the  mean,  which  is  the  property  of  arithme- 
ticals. 

The  method  of  finding  out  numbers  in  musical  proportion 
is  best  expressed  by  letters  in  Algebra. 


FELLOWSHIP,  OR  PARTNERSHIP. 

Fellowship  is  a  rule,  by  which  any  sum  or  quantity  may 
be  divided  into  any  number  of  parts,  which  shall  be  in  any 
given  proportion  to  one  another. 

By  this  rule  are  adjusted  the  gains  or  loss  or  charges  of 

partners 


120  ARITHMETIC. 

partners  ia  company  ;  or  the  effects  of  bankrupts,  or  lega- 
cies  in  case  of  a  deficiency  of  assets  or  effects  ;  or  the  shares 
of  prizes  ;  or  the  numbers  of  men  to  form  certain  detach- 
ments ;  or  the  division  of  waste  lands  among  a  number  of 
proprietors. 

Fellowship  is  either  Single  or  Double.  It  is  Single,  when 
the  sharer  or  portions  are  to  be  proportional  each  to  one  sin- 
gle given  number  only  ;  as  when  the  stocks  of  partners  are 
all  employed  for  the  same  time  :  And  Double,  when  each 
portion  is  to  be  proportional  to  two  or  more  numbers  ;^as 
when  the  stocks  of  partners  are  employed  for  different  times.' 


SINGLE  FELLOWSHIP. 

GENERAL  RULE. 

Add  together  the  numbers  that  denote  the  proportion  of 
the  shares.     Then  say, 

As  the  sum  of  the  said  proportional  numbers, 
Is  to  the  whole  sum  to  be  parted  or  divided, 
So  is  each  several  proportional  number. 
To  the  corresponding  share  or  part. 

Or,  as  the  whole  stock,  is  fo  the  whole  gain  or  loss, 
So  is  each  man's  particular  stock, 
To  his  particular  share  of  the  gain  or  loss. 

To  yROVE  THE  Work.  Add  all  the  shares  or  parts  to- 
getlfer,  and  the  sum  will  be  equal  to  the  whole  number  to 
be  shyed,  when  the  work  is  right. 

EXAMPLES. 

1.  To  divide  the  number  240  into  three  such  parts,  as 
shall  be  in  proportion  to  each  other  as  the  three  numbers  1 , 
2  and  3. 

Here  1  -|-  2  -f-  3  =  6,  the  sum  of  the  numbers. 
Then,  as  6  :  240  :  :  1  :    40  the  1st  part, 
and  as  6  :  240  :  :  2  :     80  the  2d  part, 
also  as  6  :  240  :  :  3  :  120  the  3d  part, 

Sum  of  all  240,  the  prooj^ 

Ex.  2. 


SINGLE  FELLOWSHIP.  121 

Ex.  2.  Three  persons,  a,  b,  c,  freighted  a  sliip  with  340  tuns 
of  wine  ;  of  which,  a  loaded  110  tuns,  b  97,  and  c  the  rest  : 
in  a  storm  the   seamen  were  obliged  to  throw  overboard  85 
tuns  ;  how  much  must  each  person  sustain  of  the  loss  ? 
Here    1 10  -f    97  =  207  tuns,  loaded  by  a  and  b  ; 
theref.  340—  207  =  133  tuns,  loaded  by  c. 
Hence,  as  340  :  86  :  :   1 10 

or  as  4  :  1  :  :  1 10  :  27i  tuns  =  a's  loss  ; 
and  as  4  :  1  :  :  97  :  24i  tuns  =  b's  loss  ; 
also  as      4  :     1    :  :   133  :  33i  tuns  =  c's  loss  ; 

Sum  85  tuns,  the  proof. 

3.  Two  merchants,  c  and  d,  made  a  stock  of  120/,  of 
which  c  contributed  75/,  and  d  the  rest  :  by  trading  they 
gained  30/  ;  what  must  each  have  of  it  ? 

Ans.  c  18/  15*,  and  d  11/  5s. 

4.  Three  merchants,  e,  f,  g,  made  a  stock  of  700/,  of 
which  E  contributed  123/,  f  358/,  and  g  the  rest  :  by  trading 
they  gain  125/  10s  j  what  must  each  have  of  it  ? 

Ans.  E  must  have  22/  Is  Od  2-^-^q, 
F  -  -  -  64  3  8  Off. 
G       -    -    -    39    6    3   ^3V. 

5.  A  General  imposing  a  contribution*  of  700/  on  four 
villages,  to  be  paid  in  proportion  to  the  number  of  inhabitants 
contained  in  each  ;  the  1st  containing  250,  the  2d  350,  the 
3d  400,  and  the  4th  600  persons  ;  what  part  must  each  vil- 
lage pay  1  Ans.  the  1st  to  pay  116/  13s  4rf. 

the  2d  -  -  163  6  8 
the  3d  -  -  186  13  4 
the  4th  -     -     233     6     8 

6.  A  piece  of  ground,  consisting  of  37  ac  2  ro  14  ps,  is 
to  be  divided  among  three  persons,  l,  m,  and  n,  in  propor- 
tion to  their  estates  :  now  if  l's  estate  be  worth  500/  a  year, 
m's  320/,  and  k's  75/  ;  what  quantity  of  land  must  each  one 
have  I  Ans.  l  must  have  2lk  ac  3  ro  39if  |ps. 

M     -     -     -     13       1       30yY^. 
N     -     -     -       3       0       23iif. 

V.^  A  person  is  indebted  to  o  57/  16s,  to  p  108/  3s  8(/, 
to  Q  22/  lOrf,  and  to  r  73/  ;  but  at  his  decease,  his  effects 


*  Contribution  is  a  taj^  paid  by  provinces,  towns,  villag-es,  &c.  to 
excuse  them  from  being  plundered.  It  is  paid  in  provisions  or  in 
money,  and  sometimes  in  both.  -' 

VoK.  r,  17  are 


122  ARITHMETIC. 

are  found  to  be  worth  no  more  than  170Z  145  ;  how  must  it 
be  divided  among  his  creditors  ? 

,  Ans.  o  must  have  Sll  16s  5d  2^5JL«L2^y, 
p     -     -     -     70  15  2    ^j\'^-^\. 
^     -     .     -     14     8  4    O/^-'^o^, 
R     -     -     -     47   14  11  2//j«^V 
Ex.  8.  A  ship  worth  900/,  being  entirely  lost,  of  which  i  be- 
longed to  s,  i  to  T,  and  the  rest  to  v  ;  what  loss  will  each 
sustain,  supposing  640Z  of  her  were  insured  ? 

Ans.  s  will  lose  46/,  t  90/,  and  v  225/. 

9.  Four  persons,  w,  x,  y,  and  z,  spent  among  them  26s, 
and  agree  that  w  shall  pa}-  ^  of  it,  x  i,  y  i,  and  z  }  ;  that 
is,  their  shares  are  to  be  in  proportion  as  ^,  ^,  i,  and  i  : 
what  are  their  shares  ?  Ans.  w  must  pay  9s  ^d  3^q. 

X  -  -  6  5  34^. 
Y  -  -  4  10  ^^, 
z       -     -      3   10    3^V- 

10.  A  detachment,  consisting  of  6  companies,  being  sent 
into  a  garrison,  in  which  the  duty  required  76  men  a  day  ; 
what  number  of  men  must  be  furnished  by  each  company,  ia 
proportion  to  their  strength  ;  the  first  consisting  of  54  men, 
the  2d  of  51  men,  the  3d  of  48  men,  the  4th  of  39,  and  the 
5th  of  36  men  ? 

Ans.  The  1st  must  furnish  18,  the  2d  17,  the  3d  16,  the 
4th  13,  and  the  5th  12  men.* 


DOUBLE  FELLOWSHIP. 


Double  Fellowship,  as  has  been  said,  is  concerned  im 
cases  in  which  the  stocks  of  partners  are  employed  or  contin- 
ued for  different  times. 


*  Qiiestions  of  this  nature  frequently  occurring  in  military  service. 
General  Hsviland,  an  officer  of  great  merit,  contrived  an  ingenious  in- 
strument, for  more  expeditiously  resolving  them  ;  which  is  distinguish- 
ed by  the  name  of  the  inventor,  being  called  a  Haviland* 


Rule. 


DOUBLE  FELLOWSHIP.  123 

Rule.* — Multiply  each  person's  stock  by  the  time  of  its 
continuance  ;  then  flivide  the  quantity,  as  in  Single  Fellow* 
ship,  into  shares,  in  proportion  to  these  products,  by  saying, 

As  the  total  sum  of  all  the  said  products,  I 

Is  to  the  whole  gain  or  loss,  or  quantity  to  be  parted, 

So  is  each  particular  product, 

To  the  corre^ondent  share  of  the  gain  or  loss. 


EXAMPLES. 

1.  A  had  in  company  60Z  for  4  months,  and  6  had  601  for 
B  months  ;    at  the  end  of  which  time  they  find  24i  gained  : 
how  must  it  be  divided  between  them  ? 
Here    50         60 
4  5 

200  +  300  =  600 

Then,  as  500  :  24  :  :  200  :    9|  =    9/  12*  =  a's  share. 

and  as  500  :  24  :  :  300  :  14|  =  14     8    =  b's  share. 

2.  0  and  d  hold  a  piece  of  ground  in  common,  for  which 

they  are  to  pay  54Z.     c  put  in  23  horses  for  27  days,  and   d 

21  horses  for  39  days  ;  how  much  ought  each  man  to  pay  of 

the  rent  ?  Ans.  c  must  pay  23/  5s  9d. 

D  must  pay  30  14  3 

4.  Three  persons,  e,  f,  g,  hold  a  pasture  in  common, 
for  which  they  are  to  pay  391  per  annum  ;  into  which  e  put 
7  oxen  for  3  months,  f  put  9  oxen  for  5  ihonths,  and  g  put 
in  4  oxen  for  12  months  ;  how  much  must  each  person  pay 
of  the  rent  ?  Ans.  e  must  pay  51  10s  6d  lj%q. 

p     -      -     11    16  lU  OyV 
G     -      -     12    12     7  2yV 

4.  A  ship's  company  take  a  prize  of  1000/,  which  they 
agree  to  divide  among  them  according  to  their  pay  and  the 
time  they  have  been  on  board  :  now  the  officers  and  midship- 
men have  been  on  board  6  months,  and  the  sailors  3  months  ^ 


*  The  proof  of  this  rule  is  as  follows  :  When  the  times  are  equal 
the  shares  of  the  gain  or  loss  are  evidently  as  the  stocks,  as  in  Single 
Fellowship  ;  and  when  the  stt^cks  are  equal,  the  shares  as  the  times  ; 
therefore,  when  neither  are  equal  the  shares  must  be  as  their  products. 

the 


lU  ARITHMETIC. 

the  officers  have  40s  a  month,  the  midshipmen  30*,  and  the 
sailors  225  a  month  ;  moreover  there  are  4  officers,  12  mid- 
shipmen, and  no  sailors  ;  what  will  each  man's  share  be  ? 

Ans.  each  officer  must  have  23/  2s  5d  O^^^q. 

each  midshipman     -      17  6  9     3y\\. 

each  seaman   -     -     -     6  7  2     0y|^. 

Ex.  5.  H,  with  a  capital  of  lOOOZ,  began  trade  the  first  of 
January,  and,  meeting  with  success  in  business,  took  in  i  as  a 
partner,  with  a  capital  of  1500/,  on  the  first  of  March  fol- 
lowing. Three  months  after  that  they  admit  k  as  a  third 
partner,  who  brought  into  stock  2800/.  After  trading  toge- 
ther till  the  end  of  the  year,  they  find  there  has  been  gained 
1776/  10s;  how  must  this  be  divided  among  the  partners  ? 

Ans.  H  must  have  457/  9s  4\d. 
I    -     -     -     571   16    81. 
K  -     -     -     747     3  Hi. 

6.  X,  y,  and  z  made  a  joint-stock  for  12  months  ;  x  at 
first  put  in  20/,  and  4  months  afi;er  20/  more  ;  y  put  in  at 
first  30/,  at  the  end  of  3  months  he  put  in  20/  more,  and  2 
months  after  he  put  in  40/  more  ;  z  put  in  at  first  60/,  and 
6  months  after  he  put  in  10/  more,  1  month  after  which  he 
took  out  30/ ;  during  the  12  months  they  gained  60/ ;  how 
much  of  it  must  each  have  ? 

Ans.  X  must  have  10/  18s  6d  3f  f  y. 
Y    -     -     -     22     8    1    Oif 
z    -     -     -     16  13   4    0. 


SIMPLE  INTEREST. 


Interest  is  the  premium  or  sum  allowed  for  the  loan,  or 
forbearance  of  money.  The  money  lent,  or  forborn,  is  called 
the  Principal.  And  the  sum  of  the  principal  and  its  interest, 
avlded  together,  is  called  the  Amount.  Interest  is  allowed 
at  so  much  per  cent,  per  annum  ;  which  premium  per  cent 
per  annum,  or  interest  of  100/  for  a  year,  is  called  the  rate  of 
interest : — So, 

When 


SIMPLE  INTEREST.  125 

When  interest  is  at  3  per  cent,  the  rate  is  3  ; 

-  4  per  cent.     -         -     4  ; 

-  6  per  cent,     -         -     5  ; 

-  6  per  cent.     -         -     6  ; 

But,  by  law  in  England,  interest  ought  not  to  be  taken  higher 
than  at  the  rate  of  5  per  cent. 

Interest  is  of  two  sorts  ;  Simple  and  Compound. 

Simple  Interest  is  that  which  is  allowed  for  the  principal 
lent  or  forborn  only,  for  the  whole  time  of  forbearance. 
As  the  interest  of  any  sum,  for  any  time,  is  directly  propor- 
tional to  the  principal  sum,  and  also  to  the  time  of  continu- 
ance ;  hence  arises  the  following  general  rule  of  calcula- 
tion. 

As  100/  is  to  the  rate  of  interest,  so  is  any  given  principal  to 
it  interest  for  one  year.     And  again, 

As  1  year  is  to  any  given  time,  so  is  the  interest  for  a  year, 
just  found,  to  the' interest  of  the  given  sum  for  that  time. 

Otherwise.  Take  the  interest  of  I  pound  for  a  year, 
which  multiply  by  the  given  principal,  and  this  product  again 
by  the  time  of  loan  or  forbearance,  in  years  and  parts,  for  the 
interest  of  the  proposed  sum  for  that  time. 

JVbic,  When  there  are  certain  parts  of  years  in  the  time, 
as  quarters  or  months,  or  days  :  they  may  be  worked  for, 
either  by  taking  the  aliquot  or  like  parts  of  the  interest  of  a 
year,  or  by  the  Rule  of  Three,  in  the  usual  way.  Also  to 
divide  "by  100,  is  done  by  only  pointing  off  two  figures  for 
decimals. 

EXAMPLES. 

1.  To  find  the  interest  of  230Z  IO5,  for  1  year,  at  the  rate  6f 
4  per  cent,  per  annum. 

Here,  As  100  :  4  :  :  2S01  10s  :  91  4s  4^d, 
4 


100)  9,22  0 
20 

4'40 
12 


4-80         Ans.  9/  4s  4f  df. 
4 

3-20 


Ex.2. 


126  ARITHMETIC. 

Ex.  2.  To  find  the  interest  of  647/  15s,  for  3  years,  at  6  per 
cent,  per  annum. 

As  100  :  5  :  ;  647-75  : 
Or  20  :  1  :  :  647*75  :  27-3875  interest  for  1  year. 


Z  82-1625  ditto  for  3  years. 
20 


8  3-2600 
12 


d  3-00  Ans.  82/  5s  3d. 


3.  To  find  the  interest  of  200  guineas,  for  4  years  7  months 
and  25  days,  at  4i  per  cent,  per  annum. 

ds  I  ds 

210/  As  366  :  :  9-45  :  25  :     / 

4i  or     73  :  :  9-46  :    5  :  -647^ 

5 

840  

105  73)  47-25  (-6472 

345 

9-45  interest  for  1  yr.  630 

4  <  19 

37-80       ditto  4  years. 
6  mo  =i  4-725     ditto  6  months.       /^ 
1  mo  =i     -7875  ditto  1  month. 
•6472  ditto  25  days. 

/  43-9597 
20 

s  19-1940  ' 

12 

d    2-3280 

4  Ans.  43/  19s  2^d, 


q>  1-3120 


4.  To  find  the  interest  of  450/,  for  a  year  at  6  per  cent, 
per  annum.  Ans.  22/  10s. 

6.  To  find  the  interest  of  715/  12s  6d,  for  a  year,  at  4i  per 
cent,  per  annum.  i  Ans.  32/  4s  Od. 

6.  To  find  the  interest  of  720/>  for  3  years,  at  6  per  cent, 
per  annum.  Ans.  108/. 

^  Ex.  7. 


COMPOUND  INTEREST.  127 

7.  To  find  the  interest  of  355/  15s  for  4  years,  at  4  per 
cent,  per  annum.  Ans.  66/  18s  4^d. 

Ex.  8.  To  find  the  interest  of  32/  5s  Sd,  for  7  years,  at  4{ 
per  cent,  per  annum.  Ans.  9/  12s  \d, 

9.  To  find  the  interest  of  170/,  for  li  year,  at  5  per  cent, 
per  annum.  Ans.  12/  5s. 

10.  To  find  the  insurance  on  205/  16s,  for  i  of  a  year,  at 
4  per  cent,  per  annum.  Ans.  2/  Is  l|d. 

11.  To  find  the  interest  of  319/  6d,  for  6f  years,  at  3|  per 
cent,  per  annum.  Ans.  68/  15s  9^d. 

12.  To  find  the  insurance  on  207/,  for  117  days,  at  4^  per 
cent,  per  annum.  Ans.  1/  12s  Id. 

13.  To  find  the  interest  of  17/  6s,  for  117  days,  at  4f  per 
cent,  per  annum.  Ans.  5s  3c?. 

14.  To  find  the  insurance  on  712/  6s,  for  8  months,  at  7^ 
per  cent,  per  annum.  Ans.  35/  12s  3^d. 

Note.  The  Rules  for  Simple  Interest,  serve  also  to  calcu- 
late Insurances,  or  the  Purchase  of  Stocks,  or  any  thing  else 
that  is  rated  at  so  much  per  cent. 

See  also  more  on  the  subject  of  Interest,  with  the  algebraical 
expression  and  investigation  of  the  rules  at  the  end  of  the 
Algebra,  next  following. 


COMPOUND  INTEREST. 

Compound  Interest,  called  also  Interest  upon  Interest, 
IS  that  which  arises  from  the  principal  and  interest,  taken 
together,  as  it  becomes  due,  at  the  end  of  each  stated  time  of 
payment.  Though  it  be  not  lawful  to  lend  money  at  Com- 
pound Interest,  yet  in  purchasing  annuities,  pensions,  or  leases 
in  reversion,  it  is  usual  to  allow  Compound  Interest  to  the 
purchaser  for  his  ready  money.  . 

Rules. — 1.  F^nd  the  amount  of  the  given  principal,  for  the 
time  of  the  first  payment,  by  Simple  Interest.  Then  con- 
sider this  amount  as  a  new  principal  for  the  second  payment, 
whose  amount  calculate  as  before.  And  so  on  through  all 
the  payments  to  the  last,  always  accounting  the  last  amount 
as  a  new  principal  for  the  next  payment.  The  reason  of 
which  is  evident  from  the  definition  of  Compound  Interest. 
Or  ehe^ 

2.  Find  the  amount  of  1  pound  for  the  time  of  the  first 
payment,  and  raise  or  involve  it  to  the  power  whose  index 
is  denoted  by  the  number  of  payments.  Then  that  power 
multiplied  by  the  given  principal,  will  produce  the  whole 

amount 


m  ARITHMETIC. 

amount.  From  which  the  said  principal  being  subtracted, 
leaves  the  Compound  Interest  of  the  same.  As  is  evident 
from  the  first  Rule. 

EXAMP1.ES. 

1.  To  find  the  amount  of  720/,  for  4  years,  at  6  per  cent, 
per  annum, 

Here  6  is  the  20th  part  of  100,  and  the  interest  of  1/  for  a 
•  year  is  -^\  or  -05,  and  its  amount  1-05.     Therefore, 

1.  By  the  1st  Rule.  2.  By  the  2d  Rule. 

Is     d  1'05  amount  of  11. 

20)720     0     0     1st  yr's  princip.  1-05 

36     0     0     1st  yr's  interest. 


20)756     0     0     2d  yr's  princip.  M025 

37   16     0     2d  yr's  interest. 


M  026  2d  power  of  it. 


20)793   16     0     3d  yr's  princip.  720 

39  13     91  3d  yr's  interest. 

/875-1645 

20  )  833     9     91  4th  yr's  princip.  20 

41   13     6f  4th  yr's  interest.  

s  3-2900 


1-21530625  4th  pow.  of  it. 


£815     3     3i  the  whole  amount.  12 

or  ans.  required.  

d  3-4800 


2.  To  find  the  amount  of  60/,  in  5  years,  at  5  per  cent,  per 
annum,  compound  interest.  Ans.  631  iGs  3^d. 

3.  To  find  the  amount  of  501  in  6  years,  or  10  half-years,  at 
•5  per  cent  per  annum,  compound  interest,  the  interest  payable 
half-yearly.  -Ans.  64/  Os  Id. 

4.  To  find  the  amount  of  60/,  in  5  years,  or  20  quarters,  at 
5  per  cent,  per  annum,  compound  interest,  the  interest  pay- 
able quarterly.  Ans.  64/  2s  O^d. 

5.  To  find  the  compound  interest  of  370/  forborn  for  6 
years,  at  4  per  cent,  per  annum.  Ans.  98/  3s  4}d. 

6.  To  find  the  compound  interest  of  410/  forborn  for  2i 
years,  at  4^  per  cent,  per  annum,  the  interest  payable  half- 
yearly.        "  Ans.  48/ 46  11  ic^. 

7.  To  find  the  amount,  at  compound  interest,  of  2 17/,  for- 
born for  21  years,  at  5  p^r  cent  per  annum,  the  interest  pay- 
able quarterly.  Ans.  242/  13s  41c/. 

Note.  See  the  Rules  for  Compound  Interest  algebraically 
investigated,  at  the  end  of  the  Algebra. 

ALI.IGATION, 


ALLIGATION.  12^ 


ALLIGATION, 


Alligation  teaches  how  to  compound  or  mix  together 
several  simples  of  different  qualities,  so  that  the  composition 
may  be  of  some  intermediate  quality  or  rate.  It  is  com- 
monly distinguished  into  two  cases,  Alligation  Medial,  and 
Alligation  Alternate. 


ALLIGATION  MEDIAL. 

Alligation  Medial  is  the  method  of  finding  the  rate  or 
quality  of  the  composition,  from  having  the  quantities  and 
rates  or  qualities  of  the  several  simples  given.  And  it  is  thus 
performed  : 

*  Multiply  the  quantity  of  each  ingredient  by  its  rate  or 
quality  ;  then  add  all  the  products  together,  and  add  also  all 


*  Demonstration.    The  mle  is  thus  proved  by  Algebra 

Let  a,  b,  c  be  the  quantities  of  the  ingredients, 
and  «i,  w,  p  their  rates,  or  qualities,  or  prices  ; 
then  am,  bn^  cp  are  their  several  values, 
and  am  -f-  6n  4-  <:p  the  sum  of  their  vahies, 
also  a  -^  b  ■{•  CIS  the  sum  of  the  quantities, 
and  if  r  denote  the  rate  of. the  whole  composition, 

then  «  -f-  6  -f  c  X  ''  will  be  the  value  of  the  whole, 

conseq.  a  +  6  -f-  c  X  ^  ==•  aw  +  A/i  -f-  c/&, 

and  r  z=  am  ■{-  bn  +  cp  -r-  a  +  b  +  c,  which  is  the  Rule. 

Kote,  If  an  ounce  or  any  other  quantity  of  pure  gold  be  reduced  in- 
to 24  equal  parts,  these  parts  are  catled  Caracts  j  but  ,^old  is  often 
mixed  with  some  base  metal,  which  is  called  the  Alloy,  and  the  mix- 
ture 15  said  to  be  of  so  many  caracts  fine,  according  to  the  proportion 
of  pure  gold  contained  in  it ;  thus,  if  22  caracts  of  pure  gold,  and  2  of 
alloy  be  mixed  together,  it  is  said  to  be  22  caracts  fine. 

If  any  one  of  the  s-mples  be  of  little  or  no  value  with  respect  to  the 
rest,  its  rate  is  supposed  to  be  nothing  ;  as  water  mixed  with  wine, 
and  alloy  with  gold  and  silver. 

Vol.  L  18  tlic 


130  ARITHMETIC. 

the  quantities  together  into  another  sum  ;  then  divide  the 
former  sum  by  the  latter,  that  is,  the  sum  of  the  products 
by  the  sum  of  the  quantities,  and  the  quotient  will  be  the 
rate  or  quality  of  the  composition  required. 


EXAMPLES. 


1.  If  three  sorts  of  gunpowder  be  mixed  together,  viz. 
601b  at  12rf  a  pound,  44lb  at  9d!,  and  261b  at  td  a  pound  ;  how 
much  a  pound  is  the  composition  worth  ? 

Here  50,  44,  26  are  the  quantities, 
and     12,    9,    8  the  rates  or  qualities  ; 
then  ^60  X  12  =  600 

•44  X     9  =  396 

26  X     8  =  208 

120)  1204         (10^f^=10J^. 

Ans.  The  rate  or  price  is  10  ^^d  the  pound. 

2.  A  composition  being  made  of  51b  of  tea  at  Is  per  lb, 
91^  at  8s  6d  per  lb,  and  I4ilb  at  5s  lOd  per  lb  ;  what  is  a 
lb  of  it  worth  ?  Ans.  6s  iO^d. 

3.  Mixed  4  gallons  of  wine  at  4s  lOd  per  gall,  with  7  gal- 
lons at  5s  3d  per  gall,  and  9f  gallons  at  5s  8d  per  gall  ; 
what  is  a  gallon  of  this  composition  worth  ?  Ans.  5s  4^^. 

4.  A  mealman  would  mix  3  bushels  of  flour  at  3s  Bd 
per  bushel,  4  bushels  at  5s  6d  per  bushel,  and  5  bushels  at 
4s  8rf  per  bushel ;  what  is  the  worth  of  a  bushel  of  this 
mixture  ?  Ans.  4s  l^d. 

5.  A  farmer  mixes  10  busJhels  of  wheat  at  5s  the  bushel, 
with  18  bushels  of  rye  at  3s  the  bushel,  and  20  bushels  of 
barley  at  2s  per  bushel  :  how  much  is  a  bushel  of  the  mixture 
worth  ?  Ans.  3s. 

6.  Having  melted  together  7  oz  of  gold  of  22  caracts  fine, 
12i  oz  of  21  caracts  fine,  and  17  oz  of  1^  caracts  fine  :  I 
would  know  the  fineness  of  the  composition  ? 

Ans.  20  41  caracts  fine. 

7.  Of  what  fineness  is  that  composition,  which  is  made  by 
mixing  31b  of  silver  of  9  oz  fine,  with  51b  8  oz  of  10  oz 
fine,  and  lib  10  oz  of  alloy  ?  Ans.  7|i  oz  fine. 

ALLIGATION 


[131   j 


ALLIGATION  ALTERNATE. 

Alligation  Alternate  is  the  method  of  finding  what  quan- 
tity of  aiiy  number  of  simples,  whose  rates  are  given,  will 
compose  a  mixture  of  a  given  rate.  So  that  it  is  the  reverse 
of  AHigation  Medial,  and  may  be  proved  by  it. 


1.  Set  the  rates  of  the  simples  in  a  column  under  each 
other. — 2.  Connect,  or  link  with  a  continued  line,  the  rate  of 
each  simple,  which  is  less  than  that  of  the  compound,  with  one, 
or  any  number,  of  those  that  are  greater  than  the  compound  ; 
and  each  greater  rate  with  one  or  any  number  of  the  less. — 
3.  Write  t-he  difference  between  the  mixture  rate,  and  that  of 
each  of  the  simples,  opposite  the  rate  with  which  they  are 
linked. — 4.  Then  if  only  one  difference  stand  against,  any 
rate,  it  will  be  the  quantity  belonging  to  that  rate  ;  but  if  there 
be  several,  their  sura  will  be  the  quantity. 

The  examples  may  be  proved  by  the  rule  for  Alligation 
Medial. 


♦  Ikmonst.  By  connecting  the  less  rate  to  the  gre^Her,  and  placing 
the  difference  between  then)  and  the  rate  alteitiately,  the  quantities 
resulting  are  such,  that  there  is  precisely  as  much  gained  by  one 
quantity  as  is  lost  by  the  other,  and  therefoie  the  gain  and  lo&s  upon 
the  whole  is  equal,  and  is  exactly  the  pioposed  rate  :  and  the  same 
will  be  true  of  any  other  two  simples  mfin.ged  according  to  ihe  ilule 

In  like  manner,  whatever  the  number  of  simples  may  be,  and  with  - 
how  many  soever  every  one  is  linked,  since  il;  is  always  a  less  with  a 
greater  than  the  mean  price,  there  will  be  an  e'iu:il  balance  of  loss 
and  gain  betweeri  every  two,  and  consequently  an  equal  balance  on  the  . 

whole.       Q;    E.    D. 

It  is  obvious,  from  this  Rule,  that  questions  of  this  sort  admit  of  a 
great  variety  of  answers  ;  for,  having  f  i»md  one  answer ,  we  may  find 
as  many  more  as  we  please,  by  only  multiplying  or  dividing  each  of  the 
quantities  found,  by  2,  or  3,  or  4,  &c. :  the  reason  of  which  is 
evident ;  for,  if  two  quantities,  of  two  simples,  make  a  balance  of  loss 
and  gain,  with  respect  to  the  mean  price,  so  must  also  the  double  or 
treble,  the  i  or  ^  part,  or  any  other  ratio  of  these  quantities,  and  so  on 
ad  infinitum* 

These  kinds  of  questions  are  called  by  algebraists  indeterminate  or 
wn?«m<cd  problems  ;  and  by  an  analytical  process,  theorems  may  be 
raised  that  will  give  all  the  possible  answers. 


EXAMPLES. 


132  ARITHMETIC. 


EXAMPLES. 


1.  A  merchant  would  mix  wines  at  165,  at  18*,  and  at 
22s  per  gallon,  so  as  that  the  mixture  may  be  worth  205  the 
gallon  :  what  quantity  of  each  must  be  taken  ? 

r  16-^  2  at  165 

Here  20^18->^^  \2  at  18s 

-       (22^^/4  +  2=  6  at  225. 
Ans.  2  gallons  at  lGs2  gallons  at  I85,  and  6  at  22*. 

2.  How  much  wine  at  6s  per  gallon,  and  at  4s  per  gallon 
must  be  mi\ed  together,  that  the  composition  may  be  worth 
5s  per  gallon  ?  Ans.  1  qt  or'l  gall,  &c. 

3.  How  much  sugar  at  4c?,  at  6^d  and  at  llrf  per  lb,  must  be 
mixed  together,  so  that  the  composition  formed  by  them  may 
be  worth  7c?  per  lb  ? 

Ans.  1  lb,  or  1  stone,  or  1  cwt,  or  any  other  equal  quantity 
of  each  sort. 

4.  How  much  corn  at  2s  6c?,  8s  8d,  45.  and  45  8c?  per  bushel, 
must  be  mixed  together,  that  the  compound  may  be  worth 
3s  10c?  per  bushel  ? 

Ans.  2  at  2s  6c?,  2  at  3s  8rf,  3  at  45,  and  3  at  4s  8df. 

5.  A  goldsmith  has  gold  of  16,  of  18,  of  23,  and  of  24  ca- 
racts  fine  :  how  much  must  he  take  of  each,  to  make  it  21 
<!aracts  fine  ?         Ans.  3  of  16^2  of  18,  3  of  23,  and  5  of  24. 

6.  It  is  required  to  mix  brandy  at  125,  wine  at  10s,  cyder 
at  Is,  and  water  at  0  per  gallon  together,  so  that  the  mixture 
inay  be  worth  8s  per  gallon  ? 

Ans.  8  galh  of  brandy,  7  of  wine,  2  of  cyder,  and  4  of  water. 

RULE  II. 

When  the  whole  composition  is  limited  to  a  certain  quantity  : 
Find  an  answer  as  before  by  linkinjj; ;  then  say,  as  the  sum  of 
the  .quantities,  or  diiferences  thus  determined,  is  to  the  given 
quantity  ;  so  is  each  ingredient  found  by  linking,  to  the  re- 
quired quantity  of  each. 

,      EXAMPLES. 

1.  How  much  gold  of  15,  17,  18,  and  22  caracts  fine,  must 
be  mixed  together,  to  form  a  composition  of  40  oz  of  20  ca- 
racts fine  ?  .   , 

Here 


ALLIGATION  ALTERNATE.  133 


Here  20 


54.3-1-2  =  10 


16 
Then,  as  16  :  40  :  :    2:6 
and  16  :  40  .:  :  10  :  25 
Ana.  5  oz  of  15,  of  17,  and  of  18  caracts  fine,  and  25  oz  of  22 
caracts  fine*. 

Ex.  2.  A  vintner  has  wine  at  4s,  at  os,  at  5s  6rf,  and  at  6s 
a  gallon  ;  and  he  would  make  a  mixture  of  18  gallons,  so 
that  it  might  be  afforded  as  6s  4d  per  gallon  ;  how  much  of 
each  sort  must  he  take  ? 

Ans.  3  gal.  at  4s,  3  at  5s,  6  at  5s  6d,  and  6  at  6s, 


•  A  great  number  of  questions  might  be  here  given  relating  to  the 
specific  gravities  of  metals,  &c.  but  one  of  the  most  curious  may  here 
suffice. 

Hiaro,  kii%  of  Syracuse,  gave  orders  for  a  crown  to  be  made  entirely 
of  pure  gold  j  but  suspecting-  the  workmen  had  debased  it  by  mixing 
it  with  silver  ot-  copper,  he  recommended  the  discovery  of  the  fraud 
to  the  famous  Archimedes,  and  desired  to  know  the  exact  quantity  of 
alloy  in  the  crown. 

Archimedes,  in  order  to  detect  the  imposition,  procured  two  otlier 
masseS;  tlie  one  of  pure  gold,  the  other  of  silver  or  copper,  and  each 
of  the  same  weight  with  the  former  j  and  by  putting  eack%eparately 
into  a  vessel  full  of  water,  the  quantity  of  water  expelled  b/them  de- 
termined their  specific  gravities  ;  from  which,  and  their  given  weights, 
the  exact  quantities  of  gold  and  alloy  in  the  crown  may  be  determined. 

Suppose  the  weight  of  each  crown  to  bfe  lOlb,  and  that  the  water 
expelled  by  the  copper  or  silver  was  92lb,  by  the  gold  52lb,  and  by  the 
compound  crown  641b ;  what  will  be  the  quantities  of  gold  and  alloy  in 
the  crown  ? 

The  rates  of  the  simples  are  92  and  52,  and  of  the  compound  64  ; 
therefore 

g,  192-^  12  of  copper 
°*|52-^  28  of  gold 
And  the  sum  of  these  is  12  +  28  =  40,  which  should  have  been 
but  10  ;  therefore  by  the  Rule, 

40  :  10  : :  12  :  31b  of  coppcr7  ^. 

40  :  10  : :  28  :  ;^lb  of  gpld    i  ^^^  ^"^w^^' 

RULE 


134  ARITHMETIC. 


RULE  m». 

When  one  of  the  ingredients  is  limited  to  a  certain  quan- 
tity ;  Take  the  difference  between  each  price,  and  the  mean 
rate  as  before  ;  then  say,  As  the  difference  of  that  simple 
whose  quantity  is  given,  is  to  the  rest  of  the  differences  se- 
verally ;  so  is  the  quantity  given,  to  the  several  quantities  re- 
quired. 

EXAMPLES. 

1.  How  much  wine  at  6s,  at  5s  6d,  and  6s  the  gallon, 
must  be  mixed  with  3  gallons  at  4s  per  gallon,  so  that  the 
mixture  may  be  worth  5s  4d  per  gallon  ; 

l-^sN.       8+2  =  10 

,    A\)    \  8+2=  10 
Here  64    <^      rx    ) 

'^^\  yi6+4  =  2o 

>^^  16+4  =  20 
Then  10  -  10  :  :  3  :  3 
10  :  20  :  :  3  :  6 
10  :  20  :  :  3  :  6 
Ans.  3  gallons  at  5s,  6  at  5s  6d,  and  6  at  6s. 

2.  A  grocer  would  mix  teas  at  12s,  10s,  and  6s  per  lb,  with 
201b  at  4s  per  lb.  how  much  of  each  sort  must  he  take  to  make 
the  composition  worth  8s  per  lb  *? 

Ans.  201b  at  4s,  101b  at  Gs,  lOlb  at  10s  and  20lb  at  12s, 

3.  How  much  gold  of  15,  of  17,  and  of  22  caracts  fine, 
must  be  mixed  with  5  oz  of  18  caracts  fine,  so  that  the  com- 
position may  be  20  caracts  fine  ? 

Ans.  5  oz  of  15  caracts  fine,  5  oz  of  17,  and  25  of  22. 


*  In  the  very  same  manner  questions  may  be  wrought  when  several 
of  the  ingredients  are  limited  to  certain  quantities,  by  finding  first  for 
one  limit,  and  then  for  another.  The  two  last  Rules  can  need  no  de- 
monstration, as  they  evidently  result  fi-om  the  first,  the  reason  of  which 
has  been  already  explained. 


POSITION. 


SINGLE  POSITION.  135 


POSITION. 


Position  is  a  method  of  performing  certain  questions, 
which  cannot  be  resolved  by  the  common  direct  rules.  It  is 
sometimes  called  False  Position,  or  False  Supposition,  because 
it  makes  a  supposition  of  false  numbers,  to  work  with  the 
same  as  if  they  were  the  true  ones,  and  by  their  means  dis- 
covers the  true  numbers  sought.  It  is  sometimes  also  called 
Trial-and-Error,  because  it  proceeds  by  trials  of  false  num- 
bers, and  thence  finds  out  the  true  ones  by  a  comparison  of 
the  errors. — Position  is  either  Single  op  Double. 


SINGLE  POSITION. 


Single  Position  is  that  by  which  a  question  is  resolved 
by  means  of  one  supposition  only.  Questions  which  have 
their  result  proportional  to  their  suppositions  belong  to 
Single  Position  :  such  as  those  which  require  the  multipli- 
cation or  division  of  the  number  seught  by  any  proposed  num^, 
ber  ;  or  when  it  is  to  be  increased  or  diminished  by  itself,  or 
any  parts  of  itself,  a  certain  proposed  number  of  times.  The 
rule  is  as  follows  : 

Take  or  assume  any  number  for  that  which  is  required, 
and  perform  the  same  operations  with  it,  as  are  described  or 
performed  in  the  question.  Then  say.  As  the  result  of  the 
said  operation,  is  to  the  position,  or  number  assumed  ;  so  is 
the  result  in  the  question,  to  a  fourth  term,  which  will  be  the 
number  sought*. 


♦  The  reason  of  this  Rule  is  evident,  because  it  is  supposed  that  the 
results  are  proportional  to  the  suppositions- 
Thus,  rM  i  a  :  :  nz  :  Z, 

a  z 

or  —  la  I  I  —  I  z, 

n  n 

a         a  z         z 

or  —  ±  —  &c. :  a  : :  —  ±  —  S;c. :  2r, 

n        m  n        m 

and  so  on. 

EXAMPLES. 


136  ARITHMETIC, 


EXAMPLES. 

1.  A  person  after  spending  i  and  ^  of  his  money,  has  yet 
remaining  60/;  what  had  he  at  first  ? 

ijuppose  he  had  at  first  120/.  Proof. 

Now  i  of  120  is  40  i  of  144  is     48 

i  of  it  is      30  xof  144.is     36 

their  sum  is     70  their  sum         84 

which  taken  from  120  taken  from      144 

leaves    60  leaves     60  as 

Then,  50  :  120  :  :  60  :  144,  the  Answer.  per  question. 

2.  What  number  is  that,  which  being  multiplied  by  7,  and 
the  product  divided  by  6,  the  quotient  may  be  21  ?         Ads.  18. 

3.  What  number  is  that,  which  being  increased  by  i,  i, 
and  ^  of  itself,  the  sum  shall  be  75  ?  Ans.  36. 

4.  A  general,  after  sending  out  a  foraging  ^  and  i  of  his 
men,  had  yet  remftining  1000  ;  what  number  had  he  in  com- 
mand ?  Ans.  6000. 

5.  A  gentleman  distributed  52  pence  among  a  number  of 
poor  people,  consisting  of  men,  women,  and  children  :  to 
each  man  he  gave  6d,  to  each  woman,  4d,  and  to  each  child 
2d  :  moreover  there  were  twice  as  many  women  as  men,  and 
thrice  as  many  children  as  women.  How  many  were  there 
®f  each  ?  Ans.  2  men,  4  women,  and  12  children  ? 

6.  One  being  asked  his  age,  said,  iff  of  the  years  I  have 
lived,  be  multiplied  by  7,  and  |  of  them  be  added  to  the  pro- 
dust,  the  sum  will  be  219.     What  was  his  age  ? 

Ans.  45  years. 

•■11? 


DOUBLE 


[     137     ] 


DOUBLE  POSITION. 

Double  Position  is  the  method  of  resolving  certain  ques- 
tions by  means  of  two  suppositions  of  false  numbers. 

To  the  Double  Rule  of  Position  belong  such  questions  as 
ha^re  their  results  not  proportional  to  their  positions  :  such  are 
those,  in  which  the  numbers  sought,  or  their  parts,  or  their 
multiples,  are  increased  or  diminished  by  some  given  absolute 
number,  which  is  no  known  part  of  the  number  sought. 

RtJLE  I*. 

Tare  or  assume  any  two  convenient  numbers,  and  proceed 
with  each  of  them  separately,  according  to  the  conditions  of 
the  question,  as  in  Single  Position  ;  and  find  how  much  each 
result  is  diiTerent  from  the  result  mentioned  in  the  question, 
calling  these  differences  the  errors,  noting  also  whether  the 
results  are  too  great  or  too  little. 


•  Demonstr.  The  Rule  is  founded  on  this  supposition,  namely,  that 
the  first  error  is  to  the  second,  as  the  difference  between  the  true  and 
first  supposed  numbeis  is  to  the  difference  between  the  true  and  se- 
cond sup[)Osed  number  ;  when  that  is  not  the  case,  the  exact  answer 
to  the  question  cannot  be  found  by  this  Rule.— That  the  Rule  is  true, 
according  to  that  supposition,  may  be  thus  proved. 

Let  a  and  b  be  the  two  suppositions,  and  A  and  b  their  results, 
produced  by  similar  operation  ;  also  r  and  j  their  errors,  or  the 
differences  between  the  results  A  and  B  from  the  true  result  n  ; 
and  let  x  denote  the  number  sought,  answering  to  the  true  result  n  of 
the  question. 

Then  is  n  —  a  =  r,  and  n  —  B  —  «.  And,  according  to  the 
supposition  on  which  the  Rule  is  founded,  r  :s  :  :  x— a  :  x  —  bi 
hence,  by  multiplying  extremes"  and  meaps,  rx  —  rb  =>  sx  —  sa  ,• 
then,  by  transposition,   rx  -^   sx   *^   rb   —    sa  /    and,   by  division, 

rb-~8a 
X  =  ■     the  number    sought,   which  is    the    rule  when  the 

•    r—s 
results  are  both  too  little. 

If  the  results  be  both  too  great,  so  that  a  and  b  are  both  greater 
than  N  i  then  n  —  a  =  —  r,  and  n  —  b  =  —  »,  or  r  and  «  are  both 
negative  ;  hence  —  r  :  —  s  ::  x  —  a  :  x  '-  b^  but  —  r  :  —  j  : :  -f. r 
:  4f  *,  therefore  r  ■  s  i  :  x  ^^a:  x  ^  b  ;  and  the  rest  will  be  exactly 
as  in  the  former  case. 

But  if  one  result  A  only  be  too  little,  and  the  other  b  too  great, 
or  one  error  r  positive,  and  the  other  3  negative,  then  the  theorem  be- 

rb-{.3a 
comes  X  =  — . — ,  which  is  the  Rule  in  this  case,  or  when  the  errors 

are  unlike. 
Vol.  I.  19  Then 


131 


ARITHMETie. 


Then  multiply  each  of  the  said  errors  by  the  contrary  sup- 
position, namely,  the  first  position  by  the  second  error,  and 
the  second  position  by  the  first  error.     Then, 

If  the  errors  are  alike,  divide  the  difference  of  the  products 
by  the  difference  of  the  errors,  and  the  quotient  will  be  the 
answer. 

But  if  the  errors  are  unlike,  divide  the  sum  of  the  products 
by  the  sum  of  the  errors,  for  the  answers. 

JVbfe,  The  errors  are  said  to  be  ahke,  when  they  are  either 
both  too  great  or  both  too  little  ;  and  unlike,  when  one  is  too 
great  and  the  other  too  little. 

EXAMPLES. 

1.  What  number  is  that,  which  being  multiplied  by  6, 
the  product  increased  by  18,  and  the  sum  divided  by  9,  the 
quotient  shall  be  20  ? 

Suppose  the  two  numbers  18  and  30.     Then, 

First  Position.  Second  Position.  Proof. 

18     Suppose       -  30  27 

6     mult.  .6  6 


108 

180 

-162 

18 

add 

18 

18 

9)   126 

div. 

9) 

198 

9)   180 

14 

results 

22 

20 

20 

true  res 

• 

20 

+  6 

errors  unlike 

—2 

2d  pos. 

30 

mult. 

18  1st 

po^. 

Er-    ^2 

180 

36 

rors  jle 

36 

— -— 

, 

sum  8) 

216 

sum  of 

products. 

27 

Answer 

sought. 

RULE  II. 
Find,  by  trial,  two  numbers,  as  near  the  true  number  as 
convenient,  and  work  with  them  as  in  the  question  ;  marking 
the  errors  which  arise  from  each  of  them. 

Multiply  the  difference  of  the  two  numbers  assumed,  or 
found  by  trial,  by  one  of  the  errors,  and  divide  the  product  by 
the  difference  of  the  errors,  when  they  are  alike,  but  by  their 
sum  when  they  are  unlike. 

Add 


DOUBLE  POSITION.  139 

Add  the  quotient,  last  found,  to  the  number  belonging  to 
the  said  error,  when  that  number  is  too  little,  but  subtract  it 
when  too  great,  and  the  result  will  give  the  true  quantity 
souffht*. 


EXAMPLES. 

1.  So,  the  foregoiog  example,  worked  by  this  2d  rule  will 
be  as  follows  :  • 

30  positions  18  ;  their  dif.      12 

— 2  errors  -{-  6  ;  least  error     2 


sum  of  errors  8)  24  (3  subtr. 
from  the  position  30 

leaves  the  answer  27 


Ex*  2.  A  son  asking  his  father  how  old  he  was,  received 
this  answer  :  Your  age  is  now  one-third  of  mine  ;  but  5  years 
ago,  your  age  was  only  one-fourth  of  mine.  What  then  are 
their  two  ages  ?  •  Ans.  16  and  45. 

3.  A  workman  was  hired  for  20  days,  at  3s  per  day,  for 
every  day  he  worked  ;  but  with  this  condition,  that  for  every 
day  he  played,  he  should  forfeit  Is.  Now  it  so  happened,  that 
upon  the  whole  he  had  21  4s  to  receive.  How  many  days  did 
he  work  ?  Ans.  16 

4.  A  and  e  began  to  play  together  with  equal  sums  of  money  : 
A  first  won  20  guineas,  but  afterwards  lost  back  f  of  what  he 
then  had  ;  aft«-  which,  b  had  4  times  as  much  as  a.  What 
sum  did  each  begin  with  ?  ^  Ans.  100  guineas. 

5.  Two  persons,  a  and  b,  have  both  the  same  inconie,  a 
saves  i  of  his  ;  but  b,  by  spending  50t  per  annum  more  than  a, 
at  the  end  of  4  years  finds  himself  100/  in  debt.  What  does 
each  receive  and  spend  per  annum  ? 

Ans.  They  receive  125/  per  annum  ;  also  a  spends  100/, 
and  B  spends  150/  per  annum. 


*  Fop  since,  by  the  supposition,  r  :  s  :'.  x  —  a:x  -^6,  therefore  by 
^vision,  r  -  « :  j  ; :  d  —  a  :  ?c  —  6,  which  is  the  2d  Rule. 


PERMUTATIONS 


140  ARITHMETIC. 

PERMUTATIONS  AND  COMBINATIONS. 

Permutation  is  the  altering,  changing  or  varying  the 
position  or  order  of  things  ;  or  the  showing  how  many  diffe- 
rent ways  they  may  be  placed. — This  is  otherwise  called  Al- 
ternation, Changes,  or  Variation  ;  and  the  only  thing  to  be  re- 
garded here,  is  the  order  they  stand  in  ;  for  no  two  parcels 
are  to  have  all  their  quantities  placed  in  the  same  situation  : 
as,  how  many  changes  may  be  rung  on  a  number  of  bells,  or 
how  man}'  different  ways  any  number  of  persons  may  be 
placed,  or  how  many  several  variations  may  be  made  of  any 
number  of  letters,  or  any  other  things  proposed  to  be  varied. 

Combination  is  the  showing  how  often  a  less  number  of 
things  can  be  taken  out  of  a  greater,  and  combined  together, 
without  considering  their  places,  or  the  order  they  stand  in. 
This  is  somtjtimes  called  Election  or  Choice  ;  and  here  every 
parcel  must  be  different  from  all  the  rest,  and  no  two  are  to 
have  precisely  the  same  quantities  or  things. 

Combinations  of  the  same  Form,  are  those  in  which  there  are 
the  same  number  of  quantities,  and  the  same  repetitions  :  thus, 
aabc,  bbcdf  ccde^  are  of  the  same  form  ;  aabc,  abbb\  aabb,  are 
of  different  forms. 

Composition  of  Quantities,  is  the  taking  a  given  number  of 
quantities  out  of  as  many  equal  rows  of  different  quantities, 
one  out  of  every  row,  and  combining  them  together. 

Some  illustration  of  these  definitions  are  in  the  following 
Problems  : 

PROBLEM  r.  ^ 

7^0  assig)i  the  number  of  Permutations,  or  Changei,  that  can  be 
7nade  of  any  Given  Number  of  things, all  different  from  each 
other. 

EULE*. 

Multiply  all  the  terms  of  the  natural  series  of  numbers^ 
from  1  up  to  the  given  number,  continually  together,  and  the 
last  product  will  be  the  answer  required. 

EXAMPLES. 


♦  The  reason  of  the  Rule  may  be  shovm  thus  ;  any  one  thlog  a  is 
capable  only  of  one  position,  as  a* 

Any  two  things  a  and  b,  are  only  capable  of  two  variations  >  as  a^, 
6a  i  whose  number  is  expressed  by  1  X  2' 


I 
PERMUTATIONS  AND  COMBINATIONS.        141 

EXAMPLES. 

1.  How  many  changes  may  be  rung  on  6  bells? 

1 
2 

2 

6 
4 

24 

120 
6 

720  the  Answer. 
Or  1  X  2  X  3  X  4  X  6  X  6  =  720  the  Answer. 

2.  How  many  days  can  7  persons  be  placed  in  a  .different 
position  at  dinner  ?  Ans.  5040  days. 

3.  How  many  changes  may  be  rung  on  12  bells,  and  what 
time  would  it  require,  supposing  10  changes  to  be  rung  in  1 
minute,  and  the  year  to  consist  of  366  days,  6  hours,  and  49 
minutes  ? 

Ans.  479001600  changes,  and  91  years,  26  days,  22  hours, 
41  minutes. 

4.  How  many  changes  may  be  made  of  the  words  in  the 
following  verse  :  Tot  tibi  sunt  dotes,  virgo,  quot  sidera  ccelo  ? 

Ans.  40320  changes,. 


If  there  be  three  things,  a,  b,  and  c  /  then  any  two  of  them,  leaving; 
out  the  3d,  will  have  1X2  variations  j  and  consequently  when  the  3d 
is  taken  in,  there  will  be  1  X  2  X  3  variations. 

In  the  same  manner,  when  there  are  4  things,  every  three>  leaving 
out  the  4th,  will  have  i  X  '^  X  3  variations  ;  consequently  by  taking 
in  successively  the  4  left  out,  there  will  be  1X2X3x4  variations. 
And  s©  on  as  far  as  we  please- 


PROl- 


142  ARITHMETIC, 

PRQBLEM  II. 

Any  Numher  of  different  Things  being  given  ;  to  find  how  many 
Changes  can  he  made  out  of  them,  by  taking  a  Given  Kum- 
ber  of  Quantities  at  a  Time. 

RULE.* 
Take  a  series  of  numbeis,    beginning  at  the  number  of 
things  given,  and  decreasing  by   1   tc  the  number  of  qoanti- 
ties  to  be  taken  at  a  time,  and  the  product  of  all  the  terms 
will  be  the  answer  required. 

EXAMPLES. 
1.  How  many  changes  may  be  rung  with  3  bells  out  gf  8  ? 
8 

7 

66 
6 

336  the  Answer. 

Qr,  8  X  7  X  6  (=  3  terms)  =  336  the  Answer. 
!2.  How  many  words  can  be  made  with  5  letters  of  the 
alphabet,  supposing  24  letters  in  all,  and  that  a  number  of  con- 
sonants alone  will  make  a  word.  Ans,  5100480. 
3.  How  many  words  can  be  made  with  5  letters  of  the  alpha- 
bet in  each  word,  there  being  26  letters  in  all,  and  6  vowels, 
admitting  that  a  number  of  consonants  alone  will  not  make  a 
word  ?                                                                   Ans.  137858400. 

FROB- 


*  This  Rule,  expressed  in  algebraic  termsj  is  as  follows  ; 

m  X  wi  —  l  X  M— 2  X  w— 3  &c.  to  n  terms :  where  m  ;=  the  num- 
ber of  things  given,  and  n  =>  the  quantities  to  be  taken  at  a  time. 

In  order  to  demonstrate  the  Rule,  it  will  be  proper  to  premise  the 
following  Lemma ; 

Lemma.  The  number  of  changes  of  m  ihing.e,  taken  n  at  a  time,  Is  . 
equal  to  m  changes  of  m— 1  things,  tiken  n  —  1  at  a  lime. 

DemonstV'  Let  any  five  quantities  a  h  c  d  ehe  given 

First,  leave  out  the  a,  and  let  v  *•  the  number  of  all  the  variations 
of  every  two,  6c,  bd^  Sic.  that  can  be  taken  out  of  the  four  remaining 
quantities  bade. 

Now,  let  a  be  pat  in  the  first  place  of  each  of  them,  a,  6,  c,  a,  *,  d,  8m:. 
and  the  number  of  changes  which  still  remain  the  same  ;  that  i??,  v  -«= 
the  number  of  variations  of  every  3  out  of  the  5,  a,  6,  r,  </,  <?,  when  a 
is  first. 

In  like  manner,  if  b,  c,  rf.  e  be  successively  left  out,  the  number  of 
vai:iations  of  all  the  two's  will  also  be  ^  f  ;  and  putting  t,  c,  d,  e  re- 
spectively 


PERMUTATIONS  AND  COMBINATIONS.         143 


proALem  iir. 

Any  Numher  of  Things  being  given ;  of  •which  there  are  several 
given  Thiii.gs  of  one  Sort^  and  several  of  another,  ^c. ;  to 
Find  how  tnany  Changes  can  he  made  out  of  them  all. 

BULB*. 

Take  the  series  1  X  2  X  3  X  4,  &;c.  up  to  the  number 
of  things  given,  aod  find  the  product  of  all  the  terms. 

Take  the  series  1  X  2  X  3  X  4,  Sic.  up  to  the  number  of 
given  things  of  the  first  sort,  and  the  series  1  X  2  X  3  X  4, 
&c.  up  to  the  number  of  given  things  of  the  second  sort,  &c. 

Divide 


spectively  in  the  first  place,  to  make  3  quantities  out  of  5,  there  will 
still  be  V  variations,  as  before. 

But  these  are  all  the  variations  that  can  happen  of  3  things  out  of 
5,  when  a,  b.  c,  dt  e,  are  successively  put  first  j  and  therefore  the  suifta 
of  all  these  is  the  sum  of  all  the  changes  of  3  things  out  of  5. 

But  :he  sum  of  these  is  so  many  times  t»,  as  is  the  number  of  things  j 
that  IS  5t>>  or  mv,  =  all  the  changes  of  3  times  out  of  5. 

And  the  same  way  of  reasoning  may  be  applied  to  any  numbers 
whatever. 

Demon,  of  the  Rule.  Let  any  7  things,  ab  cdef  g^he.  given,  and  let 
3  be  the  number  of  quantities  to  be  taken,. 

Then  m  z=>7^  and  n  =3  3. 

Now,  It  is  evident,  that  the  number  of  changes  that  can  be  made  by 
taking  1  by  1  out  of  5  things,  will  be  5,  which  let  =  v. 

Then,  by  the  Lemma,  when  m  =  6,  and  n  =  2,  the  number  of 
changes  will  be  =  mv  ==6X5;  which  let  be  =  t>  a  second  time. 

Again,  by  the  Lemma,  when  in  =  7  and  n  =  3,  the  number  of 
changes  h  mv  =  7  X  6  X  o  ;  that  is  wit;  =  w  X  (m-1)  X 
(m—2),  continued  to  3,  or  n  terms. 

And  the  same  may  be  shown  for  any  othfer  numbers, 

*  This  Ilule  is  expressed  in  terms  thus  : 

1  X  2  X  3  X  4  X  5,  &c.  to  m 


1  X  2  X  3,  &c.  to  p  X  1  X  2  X  3.  &c.  to  7  &c. 
where  w  =:the  number  of  things  given,  j&=  the  number  of  things 
of  the  first  sort,  g  =  the  number  of  things  of  the  second  sort  &c. 
The  Demonstration  may  be  shown  as  follows  ; 

Any  two  quantities,  <7,  b,  both  different,  admit  of  2  changes  ;  but  If 
the  quantities  are  the  same,  or  a  b  becomes  a  o,  there  will  be  only 

1X2 
one  position  ;  which  may  be  expressed  by  ■    ■  =  1. 

1  X  2 
Any  3  quantities,  a,  b,  c,  all  different  from  each  other,  afford  6  varia- 
tions; but  if  the  quantities  be  all  alike,  ova  be  becomes  a  a  a,  then 

the 


144  ARITHMETIC. 

piride  the  product  of  all  the  terms  of  the  first  series  by  the 
joint  product  of  all  the  terms  of  the  remaining  ones,  and  the 
i|Uotient  will  be  the  answer  required. 
EXAMPLES. 

1.  How  many  variations  can  be  made  of  the  letters  in  the 
word  Bacchanalia  ? 

1  X  2  (=  number  of  c's)  =  2 
1  X  2  X  3  X  4  (=  number  of  a's)  =  24 
1X2X3X4X5X6X7X8X9X10X11 
(  =  number  of  letters  in  the  word)  =  39916800 
2  X  24  =  48)  39916800  (831600  the  Answer. 
151 
76 
288 

2.  How  many  different  numbers  can  be  made  of  the  follow- 
ing figures,  1220005555  ?  Ans.  12600. 

3.  How  many  varieties  will  take  place  in  the  successior.  of 
the  following  musical  notes,  fa,  fa,  fa,  sol,  sol,  la,  mi,  fa  ? 

Ans.  3360. 

the  6  variations  will  be  reduced  to  1  j  which  may  be  expressed  by 
1X2X3 

— =  1.  Again,  if  two  of  the  quantities  only  are  alike,  or  a  ^  c 

1X2x3 

becomes,  aac  ;  then  the  6  variations  will  be  reduced  to  these  3j  a  a  c, 

1X2X3 
c  aa^  and  a  c  a  ;  which  may  be  expressed  by-        — — —  ==  3. 

1X2 
Any  4  quantities,  abed,  all  different  from  each  other,  will  admit 
of  24  variations.     But  if  the  quantities  be  the  same,   or  a  b  c  d  be- 
comes a  a  a  Of  the  number  of  variations  will  be  reduced  to  one  ; 
1X2X3X4 

which  is  =s     ■  ■■         — —  =  1. 

1x2x3x4 
Again,  if  three  of  the  quantities  only  be  the    same,  or  a  b  c  d 
becomes  a  a    a   b,  the    number  of  variations   will  be  reduced  to 
these  Ataaabtaabayab   a  a,  and  b  a  a  a;  which  is  =: 
1x2x3x4 

1=4.  • 

1X2X3 

And  thus  it  may  be  shewn,  that  if  two  of  the  quantities  be  alike, 
or  the  4  quantities  be  n  a  6  c,  the  number  of  variations  will  be  re- 

1X2X3   X  4 
duced  to  12 ;  which  may  be  expressed  by  ■  =  12- 

1X2 
And  by  reasoning  in  the  same  manner,  it  will  appear,  that  the 
number  of  changes  which  can  be  made  of  the  quantities  a  *  ^  c  c,  is 

1X2X3X4X5X6 
equal  to  60  s  which  may  be  expressed  by  ----.----—-——---—-----«----- 

1X2X1X2X3 
=  60.  And  so  on  for  any  other  quantities  whateveri 

PBOa- 


PERMUTATIONS  AND  COMBINATIONS.        146 

PROBLEM  IV. 

To  Jind  the  Changes  of  any  Criven  JSfumber  of  Things^  taking  a 
Given  Number  at  a  Time  :  in  which  there  art  several  Given 
Things  of  one  Sort^  several  of  another,  ^c. 

RULE*. 

Find  all  the  different  forms  of  combination  of  all  the  given 
thinajs,  taken  as  many  at  a  time  as  in  the  question. 

Find  the  number  of  changes  in  any  form,  and  multiply  it  by 
the  number  of  combinations  in  that  form. 

Do  the  same  for  every  distinct  form,  and  the  sum  of  all  the 
products  will  give  the  whole  number  of  changes  required. 

EXAMPLES. 

1.  How  many  alterations,  or  chanses,  can  be  made  of 
every  four  letters  out  of  these  8,  aaabbbcc  ? 

No.  of  forms.  No.  of  changes. 

a^h^a'^c^b^a,  b^c 4 

o2  62,a2c2, 62^3 6 

a^bCyb^ac.c^  ab 12 

/4  X     4  =  16 

Therefore  <3X     6=18 

^3  X   12  =  36 

70  ==  number  of  changes 
—  required. 

2.  How  many  changes  can  be  made  of  every  8  letters  out 
ef  these  10  ;  aaaabbccde  ?  Ans.  22260. 

3.  How  many  different  numbers  can  be  made  out  of  1  unit, 


*  The  reason  of  this  Rule  is  plain  from  what  has  been  shown  be- 
fore, and  the  nature  of  the  problem. 

A  Rule  for  finding  the  Number  of  Fofms. 

1.  Place  the  things  so,  that  the  greatest  indices  may  be  first,  and 
the  rest  in  brder 

2.  Begin  with  the  first  letter,  and  join  it  to  the  second)  third*  fourth, 
&c.  to  i  he  last. 

3.  Then  take  the  second  letter,  and  join  it  to  the  third,  fourtht  &c, 
to  the  last.  And  so  on,  till  they  are  entirely  exhausted*  always  remem- 
bering to  reject  such  combinations  as  have  occurred  before  ;  and  this 
wiU  give  the  combinations  of  all  the  twos. 

4.  Join  the  first  letter  to  every  one  of  the  twos,  and  the  second, 
third,  &c.  as  before  ;  and  it  will  give  the  combinations  of  all  the  threes. 

5.  Proceed  n  the  same  manner  to  get  the  combinations  of  all  the 
fours,  &c.  and  you  will  at  last  get  all  the  several  forms  of  combina- 
tions, and  the  number  in  each  form. 

Vol.  L  2I>  2  twos, 


14G  ARITHMETIC. 

2  twos,  3  threes,  4  fours,  and  5  fives  ;  taken  6  at  a  time  ? 

Ans,  2111, 
PROBLEM  V. 

To  find  the  JVurnber  of  Pombinations  of  any  Given  JVufnber  of 
things  all  different  from  each  other  ^  taken  any  Given  Number 
at  a  time. 

RULE.* 

Take  the  series  1,  2,  3,  4,  &c.  up  to  the  number  to  be 
taken  at  a  time,  and  find  the  product  of  all  the  terms. 

Take  a  series  of  as  many  terms,  decreasing  by  1,  from  the 
given  number,  out  of  which  the  election  ia  to  be  made,  and 
find  the  product  of  all  the  terms. 

Divide  the  last  product  by  the  former,  and  the  quotient- 
will  be  the  number  sought. 

EXAMPLES. 

1.  How  many  combinations  can  be  made  of  6  letters  out 
of  ten  ? 


*  This  Rule>  expressed  algebraically,  is, 
tn        m  —  1         m  —  2  m  —  3 

—  X  — —  X        ".       X  Oc.  to  n  terms  ;  where  m    is  the 

12  3  4 

number  of  given  quantities,  and  n  those  to  be  taken  at  a  time. 

Demonstr.  of  the  Mule.  1.  Let  the  number  of  tilings  to  be  taken  at 
a  time  be  2,  and  the  things  to  be  combined  =  m. 

Now,  when  m,  or  the  number  of  things  to  be  combined,  is  only  two, 
as  a  and  Z>,  it  is  evident  that  there  can  be  but  one  combination,  as  ab  ; 
but  if  m  be  increaseti  by  one,  or  the  letters  to  be  combined  be  3,  as  a, 
b,  c  ;  then  it  is  plain  that  the  nr.mber  of  conibinations  will  be  increas- 
ed by  2,  since  with  each  of  the  former  letters  a  and  Z>,  the  new  letter 
c  may  be  joined.  In  this  case  therefore,  it  is  evident  that  the  whole 
number  of  combinations  will  be  truly  expressed  by  1  -f-2. 

Againt  if  m  be  increased  by  one  letter  more,  or  \he  whole  number  of 
letters  be  four,  as  o,  b,  c,  d  ;  then  it  will  appear  that  the  whole  num- 
ber of  combinations  must  be  increased  by  3,  since  with  each  of  the 
preceding  letters  the  new  letter  d  may  be  combined.  Tue  combina- 
tions therefore,  in  this  case  will  be  truly  expressed  by  1+2+3. 

And  in  Ihe  same  manner  it  may  be  shown  that  the  whole  number  of 
combinations  pf  2,  in  54hings,  will  be  1+2  +3-+  4;  of  2  in  6  things, 
1  +  2  +  3  +  4  +  5  ;  and  of  2»  in  /things,  1+2  +  3-+4  +  5  + 
6,  &c.  ;  whence,  universally,  the  number  of  combinations  of  m  things, 
taken  2  by-2,  is  =  1  +  2  •+  3  +  4  +  5  +-  C,  &c.  to  (7n— 1)  terms, 
m         m  —  1 

But  the  sum  of  this  series  is  =  ~  X  ■;  which  is  the  same  as 

12 
tl^e  rule. 

2.  Let  now  the  number  of  quantities  in  each  combination  be  sup- 
posed to  be  three. 

Then 


PERMUTATIONS  AND  COMBINATIONS.         147 

1  X2X3X4X5X6(  =  the  number  to  be  taken 
at  a  time  )  =  720. 

10  X  9X8  X7X6  X6(=  same  number  frdna  10) 
=  151200. 

Then  720  )   151200  (210  the  Answer. 
1440 

720 
720 

2.  How  many  combinations  can  be  made  of  2  letters  out  of 
the  24  letters  of  the  alphabet  ?  Ans.  276. 

3.  A  general,  who  had  often  been  successful  in  war,  was 
asked  by  his  king  what  reward  he  should  confer  upon  him  for 
his  services  ;  the  general  only  desired  a  farthing  for  every  file, 
of  10  men  in  a  file,  which  he  could  make  with  a  body  of  100 
men  ;  what  is  the  amount  in  pounds  sterling  ? 

Ans.  18031572350Z95  2(;. 


Then  it  is  plain  that  when  m  ==»  3,  or  the  things  to  be  combined  arc 
<j,  6>  c,  there  cun  be  only  one  combination.  But  if  mbe  increased  by 
1,  or  the  things  to  be  combined  are  4,  as  a,  6,  c,  d,  then  will  the  num- 
ber of  combinations  be  increased  by  3  ;  since  3  is  the  number  of  com- 
binations of  2  in  all  the  preceding  letters,  a,  b,  c,  and  with  each  two  of 
these  the  new  letter  d  may  be  combined. 

The  number  of  combinations,  ih-^refore  in  this  case,  is  1  -f-  3. 

Again,  if  m  be  increased  by  one  more,  or  the  number  of  letters  be 
supposed  5  ;  then  the  former  number  of  combinations  will  be  in- 
creased by  6,  that  is,  by  all  the  combinations  of  2  in  thti  4  preceding 
letters,  at  b,  c,  d  :  since,  as  before,  with  each  two  of  these  the  new 
letter  c  may  be  combined. 

The  number  of  combinations,  therefore,  in  this  case,  is  1  -4-  3  +  6. 

"VVhence,  universally,  the  number  of  combinations  of  »a  things,  taken 
3  by  3,  is  1  +  3  4-  6  -f  10  &c.  to  m  —  2  terms. 

VI       m  —  1        m  —  2 

But  the  sum  of  this  series  is  =  —  X — — —  X— — —  ;    which  is 
12  3 

the  same  as  the  rule. 

And  the  same  thing  will  hold,  let  the  number  of  things  to  be  taken 
at  a  time  be  what  it  will  ;  therefore  tlie  number  of  combinations  of  m 
things,  taken  n  at  a  time,  will  be  =*•  ^ 

w        m  — 1        7n— 2        wj  — 3 
•r-  X  — —  X  X  —— -,  &c.  to  n  termS'     a*  e.  d. 

1        2  r.  4 


I'RQBv 


148  ARITHMETIC. 

PROBLEM  VI. 

To  find  the  Number  of  Combinations  of  any  Given  Number  of 
Things^  by  taking  any  Given  Number  at  a  time  ;  in  which  there 
are  several  Things  of  one  Sort,  several  of  another,  ^c. 

RULE. 

Find,  by  trial,  the  number  of  different  forms  which  the 
things  to  be  taken  at  a  time  will  admit  of,  and  the  number  of 
combinations  there  are  in  each. 

Add  all  the  combinations,  thus  found  together,  and  the  sum 
will  t>e  the  number  required.  i 

EXAMPLES. 

1 .  Let  the  things  proposed  be  a  a  a  b  b  c  ;  it  is  required  to 
find  the  number  of  combinations  made  of  every  3  of  these 
quantities  ? 

Forms.  Combinations.. 

a3 1 

fl26,  a^c,  b^a,  b^c  -         -         -         -         -     4 
ab  c      -         --         -         -         -         -J 

Number  of  combinations  required  =  6 

2.  Let  aaabbbcche  proposed  ;  it  is  required  to  find  the 
number  of  combinations  of  these  quantities,  taken  4  at  a  time  ? 

Ans.  10. 

3.  How  many  combinations  are  there  in  a  a  a  a  b  b  c  c  d  e, 
taking  8  at  a  time  ?  Aps.  13. 

4.  How  many  coipbinations  are  there  in  a  a  a  aa  bb  b  b  h 
ccccddddeeee  fff  gy  taking  10  at  a  time  ?    Ans,  2619. 

PROBLEM  yiL 

To  find  the  Compositions  of  any  Number,  in  an  equal  Number  of 
^ets,  the  things  themselves  being  all  different, 

RULE*. 

Multiply  the  number  of  things  in  every  set  continually  to- 
gether, and  the  product  will  be  the  answer  required. 


♦  Demonatr.  Suppose  there  are  only  two  sets  ;  then,  it  is  plain,  that 
every  quantity  of  the  one  set  being  combined  with  every  quantity  of  the 
^Ithej",  will  make  all  the  compositions,  of  two  things  in  these  two  sets; 

and 


PERMUTATIONS  AND  COMBINATIONS.        14^ 


EXAMPLE 

1.  Suppose  there  are  four  companies,  in  each  of  which 
there  are  9  men  ;  it  is  required  to  find  how  many  ways  9 
men  may  be  chosen,  one  out  of  each  company  ? 

9 

9 


6561  the  Answer. 
Op,  9  X  9  X  9  X  9  =  6561  the  Answer. 

2.  Suppose  there  are  4  companies  ;  in  one  of  which  there 
are  6  men,  in  another  8,  and  in  each  of  the  other  two  9  ;  what 
are  the  choices,  by  a  composition  of  4  men,  one  out  of  each 
company  ?  Ans.  3888. 

3.  How  many  changes  are  there  in  throwing  6  dice  ? 

Ans.  7776. 


and  the  number  of  these  compositions  is  evidently  the  product  of  the 
number  of  quantities  in  one  set  by  that  in  the  other. 

AgaiTif  suppose  there  are  three  sets  ;  then  the  composition  of  two, 
in  any  two  of  the  sets,  being  combined  with  every  quantity  of  the  third, 
will  make  all  the  compositions  of  three  in  the  three  sets.  That  is,  the 
coMtpositions  of  two  in  any  two  of  the  sets,  being  multiplied  by  the 
number  >f  quantities  in  the  remaining'  set,  will  produce  the  composi- 
tions r)f  three  in  the  three  sets  ;  which  is  evidently  the  continual  pro- 
duct of  all  the  three  numb*  rs  in  the  three  sets. 

And  the  same  manner  of  reasoning  will  hold,  let  the  number  of  sets 
be  what  it  will.     q.  e.  d. 

Ihe  doctrine  of  permutations,  combinations,  &c.  is  of  very  exten- 
siv:-  use  in  different  parts  of  the  Mathematics  ;  particularly  in  the 
calculat.on  of  annuities  and  chances.  The  subject  might  have  been 
pursued  to  a  much  greater  length  ;  but  what  is  here  done,  will  be 
found  sufficient  for  most  of  the  purposes  to  which  things  of  this  nature 
are  applicable* 


PRACTICAL 


150  ARITHMETIC. 


PRACTICAL  QUESTIONS  iN  ARITHMETIC. 

Quest.  1.  The  swiftest  velocity  of  a  cannon-ball,  is  about 
2000  feet  in  a  second  of  time.  Then  in  what  time,  at  that 
rate,  would  such  a  ball  be  in  moving  from  the  eartlf  to  the 
sun,  admitting  the  distance  to  be  100  millions  of  miles,  and 
the  year  to  contain  365  days  6  hours. 

Ans.  SyV/A  years. 

Quest.  2.  What  is  the  ratio  of  the  velocity  of  light  to  that 
of  a  cannon-ball,  which  issues  from  the  gun  with  a  velocity  of 
1500  feet  per  second  ;  light  passing  from  the  sun  to  the  earth 
in  1}  minutes  ?  Ans.  the  ratio  of  782222|  to  1. 

Quest.  3.  The  slow  or  parade-step  being  70  paces  per 
minute,  at  28  inches  each  pace,  it  is  required  to  determine 
at  what  rate  per  hour  that  movement  is  ?         Ans.  l|if  miles. 

Quest.  4.  The  quick-time  or  step,  in  marching,  being  2 
paces  per  second,  or  120  per  minute,  at  28  inches  each  ;  then 
at  what  rate  per  hour  does  a  troop  march  on  a  route,  and  how 
long  will  they  be  in  arriving  at  a  garrison  20  miles  distant,  al- 
lowing a  halt  of  one  liour  by  the  way  to  refresh  ? 

^       (  the  rate  is  3/y  miles  an  hour. 
°**  ^  and  the  time  T^-  hr.  or  7  h.  17i  min. 

Quest.  5.  A  wall  was  to  be  built  700  yards  long  in  29  days. 
Now,  after  12  men  had  been  employed  on  it  for  11  days,  it 
was  found  that  they  had  completed  only  220  yards  of  the  wall. 
It  is  required  then  to  determine  how  many  men  must  be  added 
to  the  former,  that  the  whole  number  of  them  may  jusf  finish 
the  wall  in  the  time  proposed,  at  the  same  rate  of  working  ? 

Ans.  4  men  to  be  added. 

Quest.  6.  To  determine  how  far  500  millions  of  guineas 
will  reach,  when  laid  down  in  a  straight  line  touching  one 
another  ;  siipposing  each  guinea  to  be  an  inch  in  diameter, 
as  it  is  very  nearly.  Ans.  71'91  miles,  728  yds.  2ft,  8  in. 

Quest.  7.  Two  persons,  a  and  b,  being  on  opposite  sides 
of  a  wood,  which  is  536  yards  about,  they  begin  to  go  round 
it,  both  the  same  way,  at  the  same  instant  of  time  ;  a  goes  at 
the  rate  of  1 1  yards  per  minute,  and  b  34  yards  in  3  minutes  ; 
and  the  question  is,  how  many  times  will  the  wood  be  gone 
round  before  the  quicker  overtake  the  slower  ? 

Ans.   17  times. 

Quest. 


PtlACTICAL  QUESTIONS.  151 

Quest.  8.  a  cab  do  a  piece  of  work  alone  in  12  days, 
and  B  alone  in  14  ;  in  what  time  will  they  both  together  per- 
form a  like  quantity  of  work  ?  Ans.  6  -^.j  days. 

Quest.  9.  A  person  who  was  possessed  of  a  |  share  of 
a  copper  mine,  sold  |  of  his  interest  in  it  for  1800/ ;  what  was 
the  reputed  value  of  the  whole  at  the  same  rate  ?     Ans.  4000/. 

Qhest.  10.  A  person  after  spending  20/  more  than  i  of 
his  yearly  income,  had  then  remaining  30/  more  than  the  half 
of  it ;  what  was  his  income  ?  Ans.  200/. 

Quest.  11.     The  hour  and  minute  hand  of  a  clock  arc 

exactly  together  at  12  o'clock  ;  when  are  they  next  together  ? 

Ans.  at  1  yV  hr  or  1  hr,  /y  min. 

Quest.  12.  If  a  gentleman  whose  annual  income  is  1500/, 
spends  20  guineas  a  week  ;  whether  will  he  save  or  run  in 
debt,  and  how  much  in  the  year  ?  Ans.  save  408/. 

Quest.  13.  A  person  bought  180  oranges  at  2  a  pennj^ 
and  180  more  at  3  a  penny  ;  after  which,  selling  them  out 
again  at  5  for  2  pence,  whether  did  he  gain  or  lose  by  the 
bargain  ?  '  Ans.  he  lost  6  pence. 

Quest.  14.  If  a  quantity  of  provisions  serves  1500  men 
12  weeks,  at  the  rate  of  20  ounces  a  day  for  each  man  ;  how 
many  men  will  the  same  provisions  maintain  for  20  weeks,  at 
the  rate  of  8  ounces  a  day  for  each  man  ?  Ans.  2250  men. 

Quest.  15.  In  the  latitude  of  London,  the  distance  round 
the  earth,  measured  on  the  parallel  of  latitude,  is  about  15550 
miles  ;  now  as  the  earth  turns  round  in  23  hours  56  minutes, 
at  what  rate  per  hour  is  the  city  of  London  carried  by  this 
motion  from  west  to  east  ?  Ans.  64&||f  miles  an  hour. 

Quest.  16.  A  father  left  his  son  a  fortune,  ^  of  which  he 
ran  through  in  8.  months  ;  |  of  the  remainder  lasted  him  12 
months  longer  ;  after  which  he  had  bare  820/  left.  What 
sum  did  the  father  bequeath  his  son  ?  Ans.  1913/65  8c/. 

Quest.  17.  If  1000  men,  besieged  in  a  town  with  pro- 
visions for  5  weeks,  allowing  each  man  16  ounces  a  day,  be 
reinforced  with  500  men  more  ;  and  supposing  that  they  can- 
not be  relieved  till  the  end  of  8  weeks,  how  many  ounces  a 
day  must  each  man  have,  that  the  provision  may  last  that 
time  ?  *  Ans.  6|  ounces. 

Quest.  13.  A  younger  brother  received  8400/,  whicli 
was  just  I  of  his  elder  brother's  fortune  :  What  was  the 
father  worth  at  his  death  ?  Ans.  19200/. 

Quest. 


132  ARITHMETIC. 

Quest.  19.  A  person,  looking  on  his  watch,  was  askedl 
what  was  the  time  of  the  day,  who  answered,  It  is  between 
6  and  6  ;  but  a  more  particular  answer  being  required,  he 
said  that  the  hour  and  minute  hands  were  then  exactly  toge- 
ther :  What  was  the  time  ?  Ans.  27  y3_  min.  past  5. 

Quest.  20.  If  20  men  can  perform  a  piece  of  work  in 
12  days,  how  many  men  will  accomphsh  another  thrice  as 
large  in  one-fifth  of  the  time  ?  Ans   300. 

Quest.  21.  A  father  devised  y\  of  his  estate  to  one  of  his 
sons,  and  /j  of  the  residue  to  another,  and  the  surplus, to  his 
relict  for  life.  The  children's  legacies  were  found  to  be 
614/  6s  Bd  diflferent  :  Then  what  money  did  he  leave  the 
widow  the  use  of?  *        Ans.  1270/  Is  d^^d. 

Quest.  22.  A  person,  making  his  will,  gave  to  one  child 
if  of  his  estate,  and  the  rest  to  another.  When  these  legacies 
came  to  be  paid  the  one  turned  out  1200/  more  than  the 
other  :  What  did  the  testator  die  worth  ?  Ans.  4000/. 

Quest.  23.  Two  persons,  a  and  b,  travel  between  London 
and  Lincoln,  distant  100  miles,  a  from  London,  and  b  from 
Lincoln,  at  the  same  instant.  After  7  hours  they  meet  on  the 
road,  when  it  appeared  that  a  had  rode  1^  miles  an  hour  more 
than  B.  At  what  rate  per  hour  then  did  each  of  the  travellers 
ride  ?  Ans.  a.  7|f ,  and  b  6^^  miles. 

Quest.  24.  Two  persons,  a  and  b,  travel  between  Lon- 
don and  Exeter,  a  leaves  Exeter  at  8  o'clock  in  the  morn- 
ing, and  walks  at  the  rate  of  3  miles  an  hour,  without  inter- 
mission  ;  and  b  sets  out  from  London  at  4  o'clock  the  S3me 
evening,  and  walks  for  Exeter  at  the  rate  of  4  miles  an  hour 
constantly.  Now,  supposing  the  distance  between  the  two 
cities  to  be_  130  miles,  whereabouts  on  the  road  will  they 
meet  ?  Ans.  69f  miles  from  f^xeter. 

Quest.  25.  Ope  hundred  eggs  being  placed  on  the 
ground  in  a  straight  line,  at  the  distance  of  a  yard  from  each 
©ther  :  How  far  will' a  person  travel  who  shall  bring  them 
©ne  by  one  to  a  basket,  which  is  placed  at  one  yard  from  the 
first  egg  ?  Ans.  10100  yards,  or  5  miles  and  1300  yds. 

Quest.  26.  The  clocks  of  Italy  go  on  to  24  hours  : 
Then  how  many  strokes  do  they  strike  in  one  complete  re- 
volution of  the  index  ?  •  Ans.  300. 

Quest.  27.  One  Sessa,  an  Indian,  having  invented  the 
game  of  chess,  shewed  it  to  bis  prince,  who  was  so  delighted 

with 


PRACTICAL  QUESTIONS.  153 

t^ith  it,  that  he  promised  him  any  rewai^d  he  should  ask  ;  on 
which  Sessa  requested  that  he  might  be  allowed  one  grain  of 
wheat  for  the  first  square  on  the  chess  board,  2  for  the  second, 
4  for  the  third,  and  so  on,  doubling  continually,  to  64,  the 
whole  number  of  squares.  Now,  supposing,  a  pint  to  contain 
7680  of  these  grains,  and  one  quarter  or  8  bushels  to  be  worth 
27s  6dy  it  is  required  to  compute  the  value  of  all  the  corn  ? 

Ans.  6460468216285/  17s  3d  ^^lq. 

Quest.  28.  A  person  increased  his  estate  annually  by 
100/  more  than  the  a  part  of  it ;  and  at  the  end  of  4  'years 
found  that  his  estate  amounted  to  10342/  3s  9d.  What  had  he 
at  first?  ^  Ans.  4000/. 

Quest.  29.  Paid  1012/  10«  for  a  principal  of  750/,  taken 
in  7  years  before  :  at  what  rate  per  cent,  per  annum  did  I  pay 
interest  ?  Ans.  5  per  cent. 

Quest.  30.  Divide  1000/  among  a,  b>  c  ;  so  as  to  give 
A  120  more,  and  b  95  less  than  c 

Ansi  A  445,  b  230,  c  325. 

Quest.  31.  A  person  being  asked  the  hour  of  the  day, 
said,  the  time  past  noon  is  equal  to  f  ths  of  the  time  till  mid- 
night.    What  was  the  time  ?  Ans.  20  min.  past  6. 

Quest.  32.  Suppose  that  I  have  fV  of  a  ship  worth  1200/  ; 
what  part  of  her  have  I  left  after  selhng  f  of  f  of  my  share, 
and  what  is  it  worth  ?  Ans.  -^W*  worth  185/. 

Quest.  33.  Part  1200  acres  of  land  among  a,  b,  c  ;  so 
that  B  may  have  100  more  than  a,  and  c  64  more  than  b. 

Ans.  A  312,  B  412,  c  476. 

Quest.  34.  What  number  is  that,  from  which  if  there  be 
taken  f  of  f,  and  to  the  remainder  be  added  y j  of  y5_,  the 
sum  will  be  10  ?  Ans.  9|f . 

Quest.  35.  There  is  a  number  which  if  multiplied  by  | 
ef  1^  of  1  ^,  will  produce  1  :  what  is  the  square  of  that  number  Z 

Ans.  IV^. 

Quest.  36.  What  length  must  be  cut  off  a  board,  8^^  inches 
broad,  to  contain  a  square  foot,  or  as  much  as  12  inches  in 
length  and  12  in  breadth  ?  Ans.  16  |f  inches. 

Quest.  37.  What  sum  of  money  will  amount  to  138/  2s 
6d,  in  15  months,  at  5  per  cent,  per  annum  simple  interest  ? 

Ans.  130/. 

Quest.  38.  A  father  divided  his  fortune  among  his  three 
sons,  A,  B,  c,  giving  a  4  as  often  as  b  3,  and  c  5  as  often  as 

Vol.  I.  21 


164  ARITHMETIC. 

B  6  ;  what  was  the  whole  legacy,  supposing  a's  share  was 
4000Z.  Ans.  9600^. 

Quest.  39.  A  young  hare  starts  40  yards  before  a  grey- 
bound,  and  is  not  perceived  by  him  till  she  has  been  up  40 
seconds  ;  she  scuds  away  at  the  rate  of  10  miles  an  honr,  and 
the  dog,  on  view,  makes  after  her  at  the  rate  of  18  ;  ho,w  long 
will  the  course  hold,  and  what  ground  will  be  run  over,  count- 
ing from  the  outsetting  of  the  dog  ? 

Ans.  eOj'gSec.  and  630  yards  run. 

Quest.  40.  Two  young  gentlemen,  without  private  for- 
tune, obtain  commissions  at  the  same  time,  and  at  the  age  of 
18.  One  thoughtlessly  spends  10/  a  year  more  than  his  pay  ; 
but,  shocked  at  the  idea  of  not  paying  his  debts,  gives  his 
creditor  a  bond  for  the  money,  at  the  end  of  every  year,  and 
also  insures  his  hfe  for  the  amonnt  ;  each  bond  costs  him  30 
shilliijgs,  besides  the  lawful  interest  of  6  per  cent,  and  to  in- 
sure his  hfe  costs  him  6  per  cent. 

The  other,  having  a  proper  pride,  is  determined  never  to 
run  in  debt ;  and,  that  he  may  assist  a  friend  in  need,  perse- 
veres in  saving  IQl  every  year,  for  which  he  obtains  an  inter- 
est of  5  per  cent,  which  interest  is  every  year  added  to  his 
savings,  and  laid  out,  so  as  to  answer  the  effect  of  compound 
interest. 

Suppose  these  two  officers  to  meet  at  the  age  of  50,  when 
each  receives  from  Government  400/  per  annum  ;  that  the 
one,  seeing  his  past  errors,  is  resolved  in  future  to  spend  no 
more  than  he  actually  has,  after  paying  the  interest  for  what 
he  owes,  and  the  insurance  on  his  life. 

The  other,  having  now  something  before  hand,  means  in 
future,  to  spend  his  full  income,  without  increasing  his  stock. 

It  is  desirable  to  know  how  much  each  has  to  spend  per 
annum,  and  what  money  the  latter  has  by  him  to  assist  the 
distressed,  or  leave  to  those  who  deserve  it  ? 

Ans.  The  reformed  officer  has  to  spend  661  19s  1| •6389c?. 
per  annum. 
The  prudent  officer  has  to  spend  437Z  12s  1  If  •4379c?. 

per  annum. 
And  the  latter  has  saved,  to  dispose  of,  762Z  19s  9-1896(1 


END  OP  THE  ARITHMETIC 


[  15g] 


OF  LOGARITHMS*. 


JLjOGARITHMS  are  made  to  facilitate  troublesome  calcu- 
lations in  numbers.  This  they  do,  because  they  perform 
multiplication  by  only  addition,  and  division  by  only  subtrac- 
tion, and  raising  of  powers  by  multiplying  the  logarithai  by 
the  index  of  the  power,  and  extracting  of  roots  by  dividing 
the  logarithm  of  the  number  by  the  index  of  the  root.  For, 
logarithms  are  numbers  so  contrived,  and  adapted  to  other 
numbers,  that  the  sums  and  differences  of  the  former  shall 
correspond  to,  and  show,  the  products}  and  quotients  of  the 
latter,  &c. 

Or,  more  generally,  logarithms  are   the  rinmerical   expo- 
nents of  ratios  ;  or  they  are  a  series   of  numbers  in  arith- 
metical 


*  The  invention  of  Logaritliras  is  due  to  Lord  Napier,  Baron  of 
Merchiston,  in  Scotland,  and  is  properly  considered  as  one  of  the 
most  useful  inventions  of  modern  times.  A.  table  of  these  numbers 
was  first  published  by  the  inventor  at  Edinbui'gh,  in  the  year  1614,  in 
a  treatise  entitled  Canon  Mirijiciim  Logarithmornm  ;  which  was  eager- 
ly received  by  all  the  learned  J hroughout  Enrope.  Mr.  Henry  Briggs, 
then  professor  of  geometry  at  Gresham  College,  soon  after  the  dis- 
covery, went  to  visit  the  noble  inventor  j  after  which,  they  jointly  un- 
dertook the  arduous  task  of  computing  new  tables  on  this  subject,  and 
reducing  them  to  a  more  convenient  form  than  that  which  was  at  first 
thought  of.  But  Lord  Napier  dying  soon  after,  the  whole  burden  fell 
upon  Mr.  Briggs,  who,  with  prodigious  laboiur  and  great  skill,  made  an 
entire  Canon,  according  to  the  new  form,  for  all  numbers  from  1  to 
20000,  and  from  900u0  to  10100,  to  14  places  of  figures,  and  published 
it  at  London,  in  the  year  1624,  in  a  treatise  entitled  Arithmetica  Loga- 
rithmica,  with  directions  for  supplying  the  intermediate  parts. 

This 


166  LOGARITHMS. 

metical  progression,  answering  to  another  series  of  numbers 
in  geometrical  progression.  i 

rp,  JO,     1,     2,     3,       4,     6,     6,  Indices,  or  logarithms, 

inus,    ^  j^     2^     ^^     g^     16,  32,  64,  Geometric  progression. 

Q        JO,     1,     2,     3,     4,       5,       6,    Indices,  or  logarithms. 
^1,     3,     9,  27,  81,  243,  729,  Geometric  progression. 


^        Jo,     1,     2,  3,  4,  6,       Indices,  or  logs. 

^^      ^1,   10,  100, 


1000,    10000,   100000,  Geom.  progress. 


Where  it  is  evident,  that   the  same  indices  serve  equally 
for  any  geometric  series  ;  and  consequently  there  may  be  an 


This  Canon  was  again  published  in  Holland  by  Adrian  Vlacq,  in  the 
year  1628,  together  with  the  Logarithms  of  all  the  numbers  which 
Mv.  Briggs  had  omitted  ;  but  he  contracted  them  down  to  10  places 
of  decimals.  Mr.  Briggs  also  computed  the  Logarithms  of  the  sines, 
tangents,  and  secants,  to  every  degree,  and  centesm,  or  100th  part  of  a 
degree,  of  the  whole  quadrant  ;  and  annexed  them  to  the  natural 
sines,  tangents,  and  secants,  which  he  had  before  computed,  to  fifteen 
places  of  figures.  These  Tables,  with  their  construction  and  use, 
■were  first  published  in  the  year  1633,  after  Mr.  Brigg*s  death,  by  Mr. 
Henry  Gellibrand,  under  the  title  of  Trigonometria  Britannica. 

Benjamin  Ursinus  also  gave  a  Table  of  Napier's  Logs,  and  of  sines, 
to  every  10  seconds.  And  Chr.  Wolf,  in  his  Mathematical  Lexicon, 
says  that  one  Van  Loser  had  computed  them  to  every  single  second, 
but  his  untimely  death  prevented  their  publication.  Many  other 
authors  have  treated  on  this  subject  ;  but  as  their  numbers  are  fie- 
quenliy  inaccurate  and  incommodiously  disposed,  they  are  now  gene- 
rally neglected.  The  Tables  in  most  repute  at  present,  are  those  of 
Gardiner  in  4to,  first  published  in  the  year  1742  ;  and  my  own  Tables 
in  8vo,  first  printed  in  the  year  1785,  where  the  Logarithms  of  ail  num- 
bers may  be  easily  found  from  1  to  lOuOOOOO  ;  and  those  of  the  sines, 
tangents,  and  secants,  to  any  degree  of  accuracy  required. 

Also,  Mr.  Michael  Taylor's  Tables  in  large  4to,  containing  the  com- 
mon logarithms,  and  the  logarithmic  sines  and  tangents  to  every  se- 
cond of  the  quadrant.  And,  in  France,  the  new  book  of  logarithms 
by  Callet  ;  the  2d  edition  of  winch,  in  1795,  has  the  tables  stil)  far- 
ther extended,  and  are  printed  with  what  are  called  stereotypes,  tlie 
types  in  each  page  being  soldered  together  into  a  solid  mass  or  block. 

Dodson's  Antilogarithmic  Canon  is  likewise  a  very  elaborate  work, 
and  used  for  finding  the  numbers  answering  to  any  given  logarithm. 

endless 


LOGARITHMS.  157 

endless  variety  of  systems  of  logarithms,  to  the  same  common 
numbers,  by  only  changing  the  second  term,  2,  3,  or  10,  &c. 
of  the  geometrical  series  of  whole  numbers  ;  and  by  interpo- 
lation the  whole  system  of  numbers  may  be  made  to  enter  the 
eometric  series,  and  receive  their  proportional  logarithms, 
whether  integers  or  decimals. 

It  is  also  apparent,  from  the  nature  of  these  series,  that  if 
any  two  indices  be  added  together,  their  sum  will  be  the  index 
of  that  number  which  is  equal  to  the  product  of  the  two  terms, 
in  the  geometric  progression,  to  which  those  indices  belong. 
Thus,  the  indices  2  and  3,  being  added  together,  make  6  ;  and 
the  numbers  4  and  8,  or  the  terms  corresponding  to  those  indi- 
ces, being  multiplied  together,  make  32,  which  is  the  number 
answering  to  the  index  o. 

In  like  manner,  if  any  one  index  be  subtracted  from  another, 
the  difference  will  be  the  index  of  that  number  which  is 
equal  to  the  quotient  of  the  two  terras  to  which  those  indi- 
ces belong.  Thus,  the  index  6,  minus  the  index  4,  is  =  2  ; 
and  the  terms  corresponding  to  those  indices  are  64  and  16, 
whose  quotient  is  =  4,  which  is  the  number  answering  to  the 
index  2. 

For  the  same  reason,  if  the  logarithm  of  any  number 
be  multiplied  by  the  index  of  its  power,  the  product  will 
be  equal  to  the  logarithm  of  that  power.  Thus,  the  index  or 
logarithm  of  4,  in  the  above  series,  is  2  ;  and  if  this  number  be 
multiplied  by  3,  the  product  will  be  =  6 ;  which  is  the  loga- 
rithm of  64,  or  the  third  power  of  4. 

And,  if  the  logarithm  of  any  number  be  divided  by  the 
index  of  its  root,  the  quotient  will  be  equal  to  the  logarithm 
of  that  root.  Thus,  the  index  or  logarithm  of  64  is  6  ;  and  if 
this  number  be  divided  by  2,  the  quotient  will  be  =  3  ;  which 
is  the  logarithm  of  8,  or  the  square  root  of  64. 

The  logarithms  most  convenient  for  practice,  are  such  as 
are  adapted  to  a  geometric  series  increasing  in  a  tenfold  pro- 
portion, as  in  the  last  of  the  above  forms  ;  and  are  those 
which  are  to  be  found,  at  present,  in  most  of  the  common 
tables  on  this  subject.  The  distinguishing  mark  of  this 
system  of  logarithms  is,  that  the  index  or  logarithm  of  10 
is  1  ;    th'at  of  100  is  2  ;  that  of  1000  is  3  ;  &c.     And,  in 

decimals. 


168  LOGARITHMS. 

decimals,  the  logarithm  of  •!  is  —  1  ;  that  of  •01  is  —  2  ;  that 
of  -001  is  —  3  ;  &c.  The  log.  of  1  being  0  in  every  system. 
Whence  it  follows,  that  the  logarithm  of  any  number  between 
1  and  10,  must  be  0  and  some  fractional  parts  ;  ?nd  that  of  a 
number  between  10  and  100,  will  be  1  and  some  fractional 
parts  ;  and  so  on,  for  any  other  number  whatever.  And 
since  the  integral  part  of  a  logarithm,  usually  called  the  Index, 
or  Characteristic,  is  always  thus  readily  found,  it  is  commonly 
omitted  in  the  tables  ;  being  left  to  be  supplied  by  the  opera- 
tor himself,  as  occasion  requires. 

Another  Definition  of  Logarithms  is,  that  the  logarithm  of 
any  number  is  the  index  of  that  power  of  some  other  num- 
ber, which  is  equal  to  the  given  number.  So,  if  there  be 
N  =  r",  then  n  is  the  log.  of  N  ;  where  n  may  be  either  posi- 
tive or  negative,  or  nothing,  and  the  root  r  any  number 
"whatever,  according  to  the  different  systems  of  logarithms. 
When  n  is  =  0,  then  N  is  =  1,  whatever  the  vahie  of  r  is ; 
which  shows,  that  the  log.  of  1  is  always  0,  in  every  system  of 
logarithms.  When  n  is  =  1,  then  N  is  =  r  ;  so  that  the 
radix  r  is  always  that  number  whose  log  is  1,  in  every  sys- 
tem. When  the  radix  r  is  =  2-718281828469,  &c.  the  in- 
dices n  are  the  hyperbolic  or  Napier's  log.  of  the  numbers  N  ; 
so  that  n  is  always  the  hyp.  log.  of  the  number  N  or  (2*718 
&c.)n  . 

But  when  the  radix  r  is  =  10,  then  the  index  n  becomes 
the  common  or  Briggs's  log.  of  the  number  N  :  so  that  the 
common  log.  of  any  number  10"  or  N,  is  n  the  index  of 
that  power  of  10  which  is  equal  to  the  said  number.  Thus 
100,  being  the  second  power  of  10  will  have  2  for  its  loga- 
rithm :  and  1000,  being  the  third  power  of  10,  will  have  3 
for  its  logarithm:  hence  also,  if  60  be  =  lO'-eosoT^  then 
is  1-69897  the  common  log.  of  50.  And,  in  general,  the  fol- 
lowing decuple  series  of  terms, 

viz.  lOS  103,  102,  101,  100^  lO-S  10-2,  10-3^  jo_4, 
oV  10000,  1000,  300,  10,  1,  -1,  -01,  -001,  -0001, 
have     4,  3,  2,       1,       0,-1,       —2,     —3,      —4, 

for  their  logarithms,  respectively.  And  from  this  scale  of 
numbers  and  logarithms,  the  same  properties  easily  follow, 
as  above  mentioned* 


PROBLEM, 


LOGARITHMS.  169 

PROBLEM. 

Po   compute   the  Logarithm  to  any   of  the  Natural  Numbers 
1,  2,  3,  4,  6,  ^c. 

RULE  I*. 

Take  the  geometric  series,  1,  10,  IGO,  1000,  10000,  &c. 
and  apply  to  it  the  arithmetic  series,  0,  1,  2,  3,  4,  &c.  as 
logarithms. — Find  a  geometric  mean  between  1  and  10,  or 
between  10  and  100,  or  any  other  two  adjacent  terms  of  the 
series,  between  which  the  number  proposed  lies. — In  like 
manner,  between  the  mean,  thus  found,  and  the  nearest  ex- 
treme, find  another  geometrical  mean  ;  and  so  on,  till  you 
arrive,  within  the  proposed  limit  of  the  number  whose  loga- 
rithm is  sought. — Find  also  as  many  arithmetical  means,  in 
the  same  order  as  you  found  the  geometrical  ones,  and  these 
will  be  the  logarithms  answering  to  the  said  geometrical 
means. 

EXAMPLE. 

Let  it  be  required  to  find  the  logarithm  of  9. 
Here  the  proposed  number  lies  between  1  and  10. 

First,  then,  the  log.  of  10  is  1,  and  the  log.  of  1  is  0  ; 
theref.  14-0-5-2=1^=  -6  is  the  arithmetical  mean, 
and  ^  f 0  X 1  =  y/  10  =  3-1622777  the  geom.  mean  ; 
hence  the  log.  of  3- 1622777  is  -5. 

"Secondly,  the  log,  of  10  is  I,  and  the  log.  of  3-1622777  is  -5  : 
theref.  1  +  '5  -=-  2  =  -75  is  the  arithmetical  mean, 
and  ^  io  X  3-1622777  =  5-6234132  is  the  geom.  mean  ; 
hence  the  log.  of  3-6234132  is  -76. 

Thirdly,  the  log,  of  10  is  1,  and  the  log.  of  5-6234132  is  -76 ; 
theref.  \  -\-  -15  -^  2  =  -875  is  the  arithmetical  mean, 
and  y/  10  X  5-6236132  =  7-4989422  the  geom.  mean  ; 
hence  the  log.  of  7-4989422  is  -875. 

Fourthly,  the  log,  of  10  is  1,  and  the  log.  of  7-4989422  is  -875} 
theref.  1  -j-  '875  -r  2  =  -9375  is  the  arithmetical  mean, 
and  ^  10  X  7-4989422  =  8-6596431  the  geom.  mean  j 
hence  the  log.  oS  8-6596431  is  -9375. 


*  The  reader  who  wishes  to  inform  himself  more  particularly  con- 
cerning the  history,  nature,  and  construction  of  Logarithms,  may  con- 
sult the  Introduction  to  my  Mathematical  Tables,  lately  published, 
where  he  will  find  his  curiosity  amply  gratified. 

Fifthly, 


160  LOGARITHMS. 

Fifthly,  the  log,  of  10  is  1,  and  the  log.  of  8-6596431  is  -9876; 
theref.  14-9375 -j- 2  =  -96875  is  the  arithmetical  mean, 
and  ^  10  X  8-6596431  =  9-3057204  the  geom.  mean  ; 
hence  the  log.  of  9-3057204  is  -96875. 

Sixthly,   the   log.    of  8-6596431  is    -9373,    and  the   log.   of 
9-3057204  is  -96875  ; 

theref.  -9375  +  -96875^  2  =  -953125  is  the  arith.  mean, 
and  ^  8-6596431  X  9-3057204  =  8-9768713  the  geo- 
metric mean  ; 
hence  the  log.  of  8-9768713  is  -953125. 

And  proceeding  in  this  manner,  after  25  extractions,  it 
will  be  found  that  the  logarithm  of  8-9999998  is  -9542425;  - 
which  may  be  taken  for  the  logarithm  of  9,  as  it  differs  so. 
little  from  it,  that  it  is  sufficiently  exact  for  all  practical  pur- 
poses. And  in  this  manner  were  the  logarithms  of  almost  all 
the  prime  numbers  at  first  computed. 


RULE  U*. 


Let  b  be  the  number  whose  logarithm  is  required  to  be 
found  ;.and  a  the  number  next  less  than  b,  so  that  b  —  a  =  1, 
the  logarithm  of  a  being  known  ;  and  let  s  denote  the  sum 
of  the  two  numbers  a  -f  6.     Then 

1.  Divide  the  constant  decimal  -8685889638  &c.  by  s,  and 
reserve  the  quotient :  divide  the  reserved  quotient  by  the 
square  of  s,  and  reserve  this  quotient :  divide  this  last  quo- 
tient also  by  the  square  of  s,  and  again  reserve  the  quotient : 
and  thus  proceed,  continually  dividing  the  last  quotient  by  the 
square  of  s,  as  long  as  division  can  be  made. 

2.  Then  write  these  quotients  orderly  under  one  another, 
the  first  uppermost,  and  divide  them  respectively  by  the  odd 
numbers,  1,  3,  5,  7,  9,  &;c.  as  long  as  division  can  be  made; 
that  is,  divide  the  first  reserved  quotient  by  1,  the  second  by 
3,  the  third  by  5,  the  fourth  by  7,  and  so  on. 

3.  Add  all  these  last  quotients  together,  and  the  sum  will 
be  the  logarithm  of  6  -7-  a ;  therefore  to  this  logarithm  add 
also  the  given  logarithm  of  the  said  next  less  number  a,  so 
will  the  last  sum  be  the  logarithm  of  the  number  6  proposed. 


*  For  tke  demonstration  of  this  rule,  see  my  Mathematical  Tables, 
p.  109.  &c. 

That 


LOGARITHMS. 


161 


That  is, 

n  111 

Log.  of  6  is  log.  a'\ X  (IH 1 1 +&c.) 

8  .  352  5s4  7s« 

where  n  denotes  the  constant  given  decimal  '8686889638  &g. 


EXAMPLES. 

Ex.  1.  Let  it  be  required  to  find  the  log.  of  number  2. 
Here  the  given  number  b  is  2,  and  the  next  less  number  a 
is  1 ,  whose  log.  is  0  ;  also  the  sum  2  -{-  1  =  3  =  s,  and  it« 
square  s^  =  9.     Then  the  operation  will  be  as  follows  : 


•868588964 

289529664 

32169962 

3574440 

397160 

44129 

4903 

645 

61 


•289629664  ( 
32169962  ( 

•289629654 

10723321 

3574440  ( 

714888 

397J60  ( 

66737 

44129  ( 

4903 

4903  ( 

446 

645  ( 

42 

61  ( 

4 

log.  off   - 

•301029995 

add  log.  1  - 

•000000000 

log.  of  2    -    -301029995 

Ex.  2.  To  eomputa  the  logarithm  of  the  number  3.. 
Here  6  =  3,  the  next  less  number  a  =:  2,  and  the  sum 
a  -f  6  =  5  =  5,  whose  square  s^  is   25,  to  divide  by  which, 
always  multiply  by  '04.     Then  the  operation  is  as  follows  : 
5  )  -868588964  1  )  -173717793  (  -173717793 

25  ) 
25  ) 
25  ) 
25  ) 
25) 


-868588964 

1) 

-1737 J  7793 

3  ) 

5) 

6948712 

277948 

7  ) 

11118 

9) 

445 

11  ) 

18 

6948712  ( 

2316237 

277948  ( 

55590 

11118  ( 

1588 

445  ( 

50 

18  C 

2 

log.  off     -  -176091260 
log.  of  2  add -301029995 

log.  of  3  sought  -477121255 


Then,  because  the  sum  of  the  logarithms  of  numbers, 
gives  the  logarithm' of  their  product  ;  and  the  difference  of 
the  logarithms,  gives  the  logarithm  of  the  quotient  of  the 

Vol.  I.  22  numbers  : 


162 


LOGARITHMS. 


numbers  ;  from  the  above  two  logarithms,  and  the  logarithm 
of  10,  which  is  1,  we  may  raise  a  great  many  logarithms,  as 
in  the  following  examples  : 


EXAMPLE  3. 
Because  ^X2  =  4,  therefore 
to  log.  2     -       -3010299951 
add  log.  2  -      ^3010299961 


sum  is  log.  4    -602059991^ 


EXAMPLE  4. 

Because  2X3  =  6,  therefore 

to  log.  2      -     -301029995 

add  log.  3.   -    -477121255 


s«m  is  log.  6    -778151250 


EXAMPLE  5. 
Because  2  =  8,  therefore 
log.  2         -       -3010299961 
mult,  by  3  3 


gives  log.  8      -903089987 


EXAMPLE  6. 
Because  3^  =  9,  iht  refore 
log.  3       -       •477r^l254y'?^ 
mult,  by  2  '2 


gives  log.  9    -954242509 


EXAMPLE  7. 
Because  lp  =  5,  therefore 
from  log.  10  1-000000000 
take  log.  2        •301029995| 


leaves  log.  6    •69897«004i 


EXAMPLE  8. 
Because  3X4  =  12,  therefore 
to  log.  3      -        -477121265 
add  log.  4     -      -602059991 


gives  log.  12     1-079181246 


And  thus,  computing,  by  this  general  rule,  the  logarithms 
to  the  other  prime  numbers,  7,  11,  13,  17,  19,  23,  &c  and 
then  using  composition  and  division,  we  may  easily  find  as 
many  logarithms  as  we  please,  or  may  speedily  examine  any 
jlojgarithm  in  the  table*. 


^  There  are,  besides  these,  many  other  ingenious  methods,  which 
later  writers  hate  discovered  for  findr;g  the  logarithms  of  numbers, 
in  a  much  easier  way  than  by  the  original  inventor  ;  but,  as  they 
cannot  Vve  understood  without  a  knowledge  of  some  of  the  higher 
b/anches  ^  the  mathematics,  it  is  thought  proper  to  omit  them,  and 
to  refer  the  reader  to  those  works  which  are  Wi  itten  expressly  on  the 
subject.  It  vould  likewise  much  exceed  tlie  limits  of  this  compen- 
dium, to  point  cut  all  the  particular  artifices  that  are  made  use  of  for 
constructing  an  entire  table  of  these  numbers  ;  but  any  information  of 
this  kind,  which  the  leamer  may  wish  to  obtain,  may  be  found  in  my 
Tables,  before  mentioned. 


Description 


I^OGARITHMS,  165 


Description  and  Vie  of  the  TABLE  of  LOGARITHMS. 

Having  explained  the  manner  oi  forming  a  table  of  the  log- 
arithms of  numbers,  greater  than  unity  ;  the  next  thaig  to 
be  done  is,  to  show  how  the  logarithms  of  fractional  quanti- 
ties may  be  found.  In  order  to  this,  it  may  be  observed, 
that  as  in  the  former  case  a  geometric  series  is  supposed  to 
increase  towards  the  left,  from  unity,  so  in  the  latter  case 
it  is  supposed  to  decrease  towards  the  right  hand,  still  be- 
ginning with  unit  ;  as  exhibited  in  the  general  description, 
page  148,  where  the  indices  being  made  negative,  still  show 
the  logarithms  to  which  they  belong.  Whence  it  appears, 
that  as  -f-  1  is  the  log.  of  10,  so  —  1  is  the  log.  of  yV  or  -1  ; 
and  as  4-  2  is  the  log.  of  100,  so  —  2  is  the  log.  of  y^^  or 
•01  :  and  so  on. 

Hence  it  appears  in  general,  that  all  numbers,  which  consist 
of  the  same  figures,  whether  they  be  integral,  or  fractional,  or 
mixed,  will  have  the  decimal  parts  of  their  logarithms  the  same» 
but  diifuring  only  in  the  in  lex,  which  will  be  more  or  less,  and 
positive  or  negative,  according  to  the  place  of  the  first  figure 
of  the  number. 

Thus,  the  logarithm  of  2651  being  3-423410,  the  log.  of 
tV>  or  To  0  J  or  r  oVo  5  ^c-  P*"^*  ^^  **  >  ^^^^  ^®  ^s  follows  : 


Numbers. 

Logarithms. 

2    6    6    1 

3-423410 

2    6    5-1 

2-423410 

2    6-51 

1-423410 

2-651 

0-423410 

•2651 

•—1423410 

0  2    6    5    1 

—2-423410 

0  2    6    5    1 

—3  -423410 

Hence  it  also  appears,  that  the  index  of  any  logarithm,  n 
always  less  by  1  than  the  number  of  integer  figures  which  the 
natural  number  consists  of  ;  or  it  is  equal  to  the  distaare  of 
the  first  figure  from  the  place  of  units,  or  first  place  of  inte- 
gers, whether  on  the  left,  or  on  the  right,  of  it  :  and  this  index 
is  constantly  to  be  placed  on  the  left  hand  side  of  the  decimal 
part  of  the  logarithm. 

When  there  are  integers  in  the  given  number,  the  index  is 
always  affirmative  ;  but  when  there  no  integers,  the  index  is 
negative,  and  is  to  be  marked  by  a  shoit  line  drawn  before  it^ 
or  else  above  it.     Thus, 

A  number  having  1,  2,  3,  4,  5,  &c.  integer  places, 

the  index  of  its  log.  is  0,  1,  2,  3,  4,  &c.  or  1  less  than  those 
places. 

And 


164  LOGARITHMS. 

And  a  decimal  fraction  having  its  first  figure  in  the 

1st,  2d,  3d,  4th,  &c.  place  of  the  decimals,  has  always 

—  1,  —  2,  —  3,  — .4,  &;c.  for  the  index  of  its  logarithm. 

It  may  also  be  observed,  that  though  the  indices  of  fractional 
quantities  are  negative,  yet  the  decimal  parts  of  their  loga- 
rithms are  always  affirmative.  And  the  negative  mark  (  —  ) 
may  be  set  either  before  the  index  or  over  it. 

1.  TO  FIND,  IN    THE  TABLE,  THE  LOGARITHM  TO  ANY 
NUMBER*. 

1.  If  the  given  Number  he  less  than  100,  or  consist  of 
only  two  figures  ;  its  log  is  immediately  found  by  inspection 
in  the  first  page- of  the  table,  which  contains  all  numbers  from 
1  to  100,  with  their  logs,  and  the  index  immediately  annexed  in 
the  next  column. 

So  the  log.  of  6  is  0-698970.     The  log.  of  23  is  1-361728.  ^ 
The  log.  of  60  is  1  -698970.     And  so  on. 

2.  ff^ihe  Number  he  more  than  100  hut  less  than  10000  ; 
that  is,  consisting  of  either  three  or  four  figures  ;  the  decimal 
part  of  the  logarithm  is  found  by  inspection  in  the  other  pages 
of  the  table,  standing  against  the  given  number,  in  this  manner; 
viz.  the  first  three  figures  of  the  given  number  in  the  first 
column  of  the  page,  and  the  fourth  figure  one  of  those  along 
the  top  line  of  it ;  then  in  the  angle  of  meeting  are  the  last 
four  figures  of  the  logarithm,  and  the  first  two  figures  of  the 
same  at  the  beginning  of  the  same  line  in  the  second  column 
of  the  page  :  to  which  is  to  be  prefixed  the  proper  index, 
which  is  always  1  less  than  the  number  of  integer  figures. 

So  the  logarithm  of  251  is  2-399674,  that  is,  the  decimal 
•399674  found  in  the  table,  with  the  index  2  prefixed,  because 
the  given  number  contains  three  integers.  And  the  log.  of 
34-09  is.  1-632627,  :that  is,  the  decimal  -632627  found  in  the 
table,  with  the  index  1  prefixed,  because  the  given  number 
contains  two  integers. 

3.  But  if  the  eiven  Number  contain  more  than  four  figures  ; 
take  out  the  logarithm  of  the  first  four  figures  by  inspection 
in  the  table,  as  before,  as  also  the  next  greater  logarithm,  sub- 
tracting the  one  logarithm  from  the  other,  as  also  their  cor- 
responding  numbers  the   one   from  the    other.      Then  say. 

As  the  difference  between  the  two  numbers, 
Is  lo  the  difference  of*their  logarithms. 
So  is  the  remaining  part  of  the  given  number, 
To  the  proportional  part  of  the  logarithm. 


♦  See  the  table  of  Loffarithms  at  the  end  of  the  2d  volume. 

Which 


LOGARITHMS.  16o 

Which  part  being  added  to  the  less  logarithm,  before  taken 
•ut,  gives  the  whole  logarithms  sought  very  nearly. 

EXAMPLE. 

To  find  the  logarithm  of  the  number  340926. 
The  log.  of  340*500,  as  before,  is  532627. 

And   log.    of  341000         -         -     ^  is  532754. 
The  diffs.  are        100  and  127 

Then  as   100  :  127  ::  26  :  33,  the  proportional  part. 

This  added  to      -     -     532627,  the  first  log. 

Gives,  with  the  index,  1-532660,  for  the  log.  of  34-0926. 

4.  If  the  number  consist  both  of  integers  and  fractions,  or 
IS  entirely  fractional  :  find  the  decimal  part  of  the  logarithm 
the  same  as  if  all  its  figures  were  integral  ;  then  this,  having 
prefixed  to  it  the  proper  index,  will  give  the  logarithm  re- 
quired. 

5.  And  if  the  given  number  be  a  proper  vulgar  fraction  ; 
subtract  the  logarithm  of  the  denominator  from  the  loga- 
rithm of  the  numerator,  and  the  remainder  will  be  the  loga- 
rithm sought  ;  which,  being  that  of  a  decimal  fraction,  must 
always  have  a  negative  index. 

6.  But  if  it  be  a  mixed  number  ;  reduce  it  to  an  improper 
fraction,  and  find  the  difference  of  the  logarithms  of  the  nu- 
merator and  denominator,  in  the  same  manner  as  before. 

EXAMPLES. 


1.  To  find  the  log.  of  f  f 
Log.  of  37        -  1-568202 

Log.  of  94         -  1-973128 


Dif.  log.  of  |i       ■-- 1-595074 


Where  the  index  1  is  negative. 


2    To  find  the  log.  of  ll^. 
First,  I7J^  =  o/.     ThenV 
Log.  of  405         -         2-607455 
Log.  of  23  -        1-361728 


Dif.  log.  of  17i|  1-245727 


n.  TO  FIND  THE  NATURAL  NUMBER  TO  ANY  GIVEN 
-      LOGARITHM. 

This  is  to  be  found  in  the  tables  by  the  reverse  method 
to  tlie  former,  namely,  by  searching  for  the  proposed  loga- 
rithm pmong  those  in  the  table,  and  taking  out  the  corres- 
ponding number  by  inspection,  in  which  the  proper  number 
of  integers  are  to  be  pointed,  off,  viz.  1  more  than  the 
index.  For,  in  finding  the  number  answering  to  any  given 
logarithm,  the  index  always  shows  how  far  the  first  figure 

must. 


166  LOGARITHMS. 

must  be  removed  from  the  place  of  units,  viz.  to  the  left  hand, 
or  integers,  when  the  index  is  affirmative  ;  but  to  the  right 
hand,  or  decimals,  when  it  is  negative. 

EXAMPLES. 

So,  the  number  to  the  log.    1  -632882  is  34-11. 

And  the  number  of  the  log.  1-632882  is  '3411. 

But  if  the  logarithm  cannot  be  exactly  found  in  the  table  ; 
take  out  the  next  greater  and  the  next  less,  subtracting  the 
one  of  these  logarithms  from  the  other,  as  also  their  natural 
numbers  the  one  from  the  other,  and  the  less  logarithm  from 
the  logarithm  proposed.     Then  say, 

As  the  difference  of  the  first  or  tabular  logarithms. 
Is  to  the  difference  of  their  natural  numbers. 
So  is  the  differ,  of  the  given  log.  and  the  least  tabular  log. 
To  their  corresponding  numeral  difference. 
Which  being  annexed  to  the  least  natural  number  above  taken, 
gives  the  natural  number  sought,  corresponding  to  the  pro- 
posed logarithm. 

EXAMPLE. 


So,   to  find  the  natural   number  answering  to  the  given 

logarithm  1-532708. 
Here  the  next  greater  and  next  less  tabular  logarithms, 
with  their  corresponding  numbers,  are  as  below  : 

Next  greater  632744  its  num.  341000  ;  given  log.  632708 
Next  less        632627  its  num.  340900  ;  next  less     632627 


Differences  127  —     100  —         81 


Then,  as  127  :  100  :  i  81  :  64  nearly,  the  numeral  differ. 
Therefore  34-0964  is  the  number  sought,  marking  off  two 
integers,  because  the  index  of  the  given  logarithm  is  1. 

Had  the  index  been  negative,  thus  1*632708,  its  corres- 
ponding number  would  have  be^n  -340966,  wholly  dc' 
cimal. 

MULTIPU- 


[  167  ] 


MULTIPLICATION  BY  LOGARITHMS. 

RULE. 

Take  out  the  logarithms  of  the  factors  from  the  table, 
then  add  them  together,  and  their  sum  will  be  the  logarithm 
of  the  product  required.  Then,  by  means  of  the  table,  take 
out  the  natural  number,  answering  to  the  sum,  for  the  pro- 
duct sought. 

Observing  to  add  what  is  to  be  carried  from  the  decimal 
part  of  the  logarithm  to  the  affirmative  index  or  indices,  or 
else  subtract  it  from  the  negative. 

Also,  adding  the  indices  together  when  they  are  of  the 
same  kind,  both  affirmative  or  both  negative  ;  but  subtracting 
the  less  from  the  greater,  when  the  one  is  affirmative  and  the 
other  negative,  and  prefixing  the  sign  of  the  greater  to  the 
remainder. 


EXA> 

.  To  Multiply  23-14  by 
5-062. 
Numbers.          Logs. 
23-14  -   1-364363 
5-062  -  0-704322 

IPLES. 

2.  To  multiply  2-58192C 
by  3-457291, 
Numbers.        Logs. 
2-581926  -  0-411944 
3-457291   -  0-538736 

)duct   117-1347     2-068685 

Prod.      8-92648     -  0-950680 

3.  To  mult.  3-902  and  597-16 
and  -0314728  all  together. 
Numbers.  ,     Logs. 
3-902       -     0-591287 
597-16    -     2-776091 
•0314728—2-497935 


Prod.  73-3333 


I  865313 


Here  the  —  2  cancels  the  2, 
and  the  1  to  carry  from  the 
decimals  is  set  down. 


4.  To  mult.  3-586,  and  2-1046, 
and  0-8372,  and  0-0294  all 
together. 

Numbers.     Logs. 
3-586    -   0-654610 
2-1046  -   0-323170 
0-8372—1-922829 
00294  —  2-468347 

Prod.     0-1057618-1-26895G 


Here  the  2  to  carry  cancels 
the  —  2,  and  there  remains  the 
— 1  to  set  down. 


DIVISION 


[168] 

DIVISION  BY  LOGARITHMS. 
RULE. 

From  the  logarithm  of  the  dividend  subtract  the  logarithm 
of  the  divisor,  and  the.  number  answering  to  the  remainder 
will  be  the  quotient  required. 

Observing  to  change  the  sign  of  the  index  of  the  divisor, 
from  affirmative  to  negative,  or  fioni  negative  to  affirmative  ; 
then  take  the  sum  of  the  indices  if  they  be  of  the  same  name, 
or  their  difference  when  of  different  signs,  with  the  sign  of 
the  greater,  for  the  index  to  the  logarithm  of  the  quotient. 

And  also,  when  1  is  borrowed,  in  the  left-hand  place  of 
the  decimal  part  of  the  logarithm,  add  it  to  the  index  of  the 
divisor  when  that  index  is  affirmative,  but  subtract  it  when 
negative  ;  then  let  the  sign  of  the  index  arising  from  hence 
fee  changed,  and  worked  with  as  before. 

EXAMPLES. 


1.  To  divide  24163  by  -4567. 

Numbers.       Logs. 
Dividend     24163  -  4-383151 
Divisor    -     4567  -  3-659631 


quot.       5-29078       0-723520 


2.  To  divide  37-149  by  623-76 

Numbers.         Logs. 
Dividend    37-149  -   1-569947 
Divisor      523-76  -  2-719132 


Quot.      -0709275—  2-850816 


3.  Divide  -06314  by  -007241 
Numbers.       Logs. 
Divid.       -06314  —  2-800305 
Divisor  -007241  — ■  3-859799 


Quot. 


8-71979     0-940506 


Here  1  carried  from  the 
decimals  to  the  —  3,  makes  it 
become — 2,  which  taken  from 
the  other  —  2,  leaves  0  re- 
maining. 


4.  To  divide  -7438  by  12-9476 

Numbers.         Logs. 
Divid.        -7438    —  1-871456 
Divisor       12-9476     1-112189 


Q.uot. 


•057447  —  2-759267 


Here  the  1  taken  from  the 
— 1,  makes  it  become  —  2,  t® 
set  down. 


Note.  As  to  the  Rule-of-Three,  or  Rule  of  Proportion,  it 
is  performed  by  adding  the  logarithms  of  the  2d  and  3d  terms, 
and  subtracting  that  o?  the  first  term  from  their  sum. 


INVOLUTION 


[  169  J 
INVOLUTION  BY  LOGARITHMS. 

RULft. 

Take  out  the  logaVithm  of  the  given  number  from  the 
table.  Multiply  the  log.  thus  found,  by  the  index  of  the 
power  proposed.  Find  the  number  answering  to  the  pro- 
duct, and  it  will  be  the  power  required. 

Note.  In  multiplying  a  logarithm  with  a  negative  index, 
by  an  affirmative  number,  the  product  will  be  negative. 
But  what  is  to  be  carried  from  the  decimal  part  of  the  loga- 
rithm, will  always  be  affirmative.  And  therefore  their  dif- 
ference will  be  the  index  of  the  product,  and  is  always  to  be 
made  of  the  same  kind  with  the  greater. 
EXAMPLES. 


,  1 .  To  square  the  number 

2-6791. 

Numb.  Log. 

Root  2-6791     -     -    0-411468 

The  index    -     -     2 


Power  6-65174 


0-822936 


2.  To  find  the  cube  of 
3-07146. 
Numb.  Log, 

Root  307146    -    -    0-487346 
The  index    -    -    3 


Power  28-9758 


1-462035 


3.  To  raise  -09163  to  the  4th 
power. 
Numb.  Log. 

Root  -09163  —2-962038 

The  index     -     -     4 


Pow.  -000070494—  5-848152 


Here  4  times  the  negative 
index  being  —  8  and  3  to  carry, 
the  difference  —  5  is  the  index 
of  the  product. 


4.  To  raise   1-0045  to  the 
365th  power. 
Numb.  Log. 

Root  1-0045     -    -     0-001950 
The  index     -    -     365 


9750 
11700 
5850 


Power  5-14932*        0  711750 


*  This  answer  5'14932  though  found  strictly  according  to  the  gene- 
ral rule,  is  not  correct  in  the  last  two  figures  32  j  nor  can  the  answers 
to  such  questions  relating  to  very  high  powers  be  generally  found  true 
to  6  places  of  figures  by  the  table  of  logarithms  in  this  work  :  if  any 
power  above  the  hundred  thousandth  were  required,  not  one  figure  of 
the  answer  found  by  the  table  of  logarithms  here  given  could  be  de- 
pended on. 

The  logarithm  of  1-0045  is  00194994108  true  to  eleven  places,  which 
multiplied  by  365  gives  -7117285  true  to  7  places,  and  the  correspond- 
ing numb  c  true  to  7  places  is  5*  149067. 

Vol.  I.  23  EVOLUTION 


[  1.70  3 

EVOLUTION  BY  LOGARITHMS. 

Take  the  log.  of  the  given  number  out  of  the  table. 
Divide  the  log.  thus  found  by  the  index  of  the  root.     Then 
the  number  answering  to  the  quotient,  will  be  the  root. 

Note.  When  the  index  of  the  logarithm,  to  be  divided,  is 
negative,  and  does  not  exactly  contain  the  divisor,  without 
some  remainder,  increase  the  index  by  such  a  number  as  will 
make  it  exactly  divisible  by  the  index,  carrying  the  units  bor- 
rowed, as  so  many  tens,  to  the  left-hand  place  of  the  decimal, 
and  then  divide  as  in  whole  numbers. 


Ex.  1.  To  find  the  square  root 
of  365 
Numb.  Log. 

Power  365  2)2-562293 

Root       19-10496     l-281146i 


Ex.  2.    To  find  the  3d  root  of 

12345. 

Numb.  Log. 

Power       12345  3)  4-091491 

Root       23-1116      1-363830^ 


Ex.  3.    To  find  the  10th  root 

of  2. 

Numb.  Log. 

Power  2     -  -     10)  0-301030 

Root  1-071773  0-030103 


Ex.  4.  To  find  the  365th  root 

of  1-045. 

Numb.  Log. 

Power  1-045    365)  0-019116 

Root     1000121      0-000052-1 


EiL.  5.  To  find  ^  -093. 
Numb.  Log. 

Power  -093     2)  —  2-968483 
Root  -304959      —  1 -4842411 

Here  the  divisor  2  is  con- 
tained exactly  once  in  the  ne- 
gative index  —  2,  and  there- 
fore the  index  of  the  quotient 
is  — L 


Ex.  6.  To  find  the  %/  -00048. 

Numb.  Log.  , 

Power  -00048  3)  —  4-681241 

Root     -0782973  —  2-893747 

Here  the  divisor  3  not  being  ex- 
actly contained  in  —  4,  it  is  augment- 
ed by  2,to  make  up  6,  in  which  the 
divisor  is  contained  just  2  times  ; 
then  the  2,  thus  borrowed,  being 
carried  to  the  decimal  figure  6, 
makes  26,  which  divided  by  3, 
gives  8,  &c. 


Ek.  7.  To  find  3-1416  X  82  X  ff. 
Ex.  8.  To  find  -02916  X  751-3  X  ^fy. 
Ex.  9.  As  7^41   :  3-58   :  :  20-46  :  ? 
Ex.  10.  As  y/  724  :  v'ff  :  :  6-927  :  ? 


ALGEBRA. 


[  171  ] 


ALGEBRA. 


DEFINITIONS  AND  NOTATION. 


l.Al 


lLGEBRA  is  the  science  of  computing  by  symbols. 
It  is  sometimes  also  called  Analysis  ;  and  is  a  general  kind  of 
arithmetic,  or  universal  way  of  computation. 

2.  In  this  science,  quantities  of  all  kinds  are  represented 
by  the  letters  of  the  alphabet.  And  the  operation  to  be  per- 
formed with  them,  as  addition  or  subtraction,  Sac.  are  denoted 
by  certain  simple  characters,  instead  of  being  expressed  by 
words  at  length. 

3.  In  algebraical  questions,  some  quantities  are  known  or 
given,  viz.  those  whose  values  are  known  :  and  others  un- 
known, or  are  to  be  found  out,  viz.  those  whose  values  are  not 
known.  The  former  of  these  are  represented  by  the  leading 
letters  of  the  alphabet,  a,  fc,  c,  d,  &c.  ;  and  the  latter,  or  un- 
known quantities,  by  the  final  letters,  Zf  y,  x,  u,  &c.  %- 

4.  The  characters  used  to  denote  the  operations,  are  chiefly 
the  following : 

-f-  signifies  addition,  dnd  is  named  vhts, 

—  signifies  subtraction,  and  is  named  minus. 

X  or  .  signifies  multiplication,  and  is  named  into. 

-f-  signifies  division,  and  is  named  hy. 

y/  signifies  the  square  root ;  %/  the  cube  root  ;  \/  th«  4tit 
root,  &c.  ;  and  ^  the  nth  root. 

:  :  :  :  signifies  proportion. 

=  signifies  equality,  and  is  named  equal  to. 

And  so  on  for  other  operations. 

Thus  a  H-  6  denotes  that  the  number  represented  by  h  is  to 
be  added  to  that  represented  by  a. 

a  —  6  denotes,  that  the  number  represented  by  6  is  to  be 
subtracted  from  that  represented  by  a. 

o  OD  6  denotes  the  difference  of  a  and  6,  when  it  is  not  known 
which  is  the  greater. 

a6,  or 


172  ALGEBRA. 

,<z6,  or  a  X  h,  or  a.h,  expre^es  the  product,  by  multiplicatioD, 
«f  the  numbers  represented  by  a  and  h. 
a 
a  -f-  6,  or  — ,  denotes,  that  the  number  represented  by  a 
b 
is  to  be  divided  by  that  which  is  expressed  by  h. 

a  :  b  :  :  c  :  d,  signifies  that  a  is  in  the  same  proportion  to  6, 
as  c  is  to  d. 

X  =  a  —  6  -{-  c  is  an  equation,  expressing  that  x  is  equal  to 
the  difference  of  a  and  6,  added  to  the  quantity  c. 

i  J. 

^  a,  or  a^,  denotes  the  square  root  of  a  ;  ^  a,  or  o^,  the 

cube  root  of  a  ;  and  ^/a^  or  a^  the  cube  root  of  the  square  of  a; 

also  ^  a,  or  a»»,  is  the  mth  root,  of  a  ;  and  "^  a"  or  a  ^  is  the 

n 

nth  power  of  the  mth  root  of  a,  or  it  is  a  to  the  ^  power. 

a2  denotes  the  square  of  a  ;  a^  the  cube  of  a ;  a*  the  fourth 
power  of  a  ;  and  a"  the  nth  power  of  a. 

a-\-b  X  c,  or  («+6)  c,  denotes  the  product  of  the  compound 
quantity  a  -{-  6  multiply  by  the  simple  quantity  c.  Using  the 
bar ,  or  the  parenthesis  (  )  as  a  vinculum,  to  connect  seve- 
ral simple  quantities  into  one  compound. 

ft-f.6 

a.-\-b  -j-a-hoT ,    expressed   like    a   fraction,    means 

a —  b 
the  quotient  of  a  +  ^  divided  by  a  —  b. 

^  ab  -\-  cd,  or  [ab  -\-  cd)^j  is  the  square  root  of  the  com- 
pound quantity  ab  +  cd.  And  c  ^  ab  -\-  cd,  or  c  {ab  -\-  cd)'^j 
denotes  the  product  of  c  into  the  square  root  of  th^  compound 
quantity  ab  -\-  cd. 

3 

a-{-b  —  c,  or  (<x  +  6  —  c)^,  denotes  the  cube,  or  third 
power,  of  the  compound  quantity  a  -\-  b  —  c. 

3a  denotes  that  the  quantity  a  is  to  be  taken  3  times,  and 
4  {a  -\-  b)  is  4  times  a  -^  b.  And  these  numbers,  3  or  4, 
showing  how  often  th^e  quantities  are  to  be  taken,  or  multiplied, 
are  called  Co-efficients. 

Also  fa;  denotes  that  a;  is  multiplied  by  |  ;  thus  |  X  a;  or 
^x 

5.  Like  Quantities,  are  those  which  consist  of  the  same  let- 
ters, and  powers.  As  a  and  3a;  or  2ab  and  4ab  ;  or  3a^bc 
and  —  ba^  be. 

6.  Unlike  Quantities,  are  those  which  consist  of  different 
letters,  or  different  powers.  As  a  and  6  ;  or  2a  and  a^  ;  or 
^al-  and  3abc. 

7.  Simple 


DEFINITIONS  AND  NOTATION.  173 

7.  Simple  Quantities,  are  those  wliich  consist  of  one  term 
only.     As  3a,  or  5ab,  or  Qabc^ . 

8.  Compound  Quantities  are  those  which  consist  of  two  or 
more  terms.     As  a  +  6,  or  2a  —  3c,  or  a  -f  26  —  3c. 

9.  And  when  the  compound  quantity  consists  of  two  terms, 
it  is  called  a  Binomial,  as  a  +  6  ;  when  of  three  terms,  it  is  a 
Trinomial,  as  a  4-  26  —  3c  ;  when  of  four  terms,  aQuadrino- 
mial,  as  2a  —  36  +  c  —  4d  ;  and  so  on.  Also,  a  Multinomial 
or  Polynomial,  consists  of  many  terms. 

10.  A  Residual  Quantity,  is  a  binomial  having  one  of  the 
terms  negative.     As  a  —  26. 

11.  Positive  or  Affirmative  Quantities,  are  those  which  are 
to  be  added,  or  have  tl^  sign  -f-  As  a  or  -}-  «>  or  a6 ;  for 
when  a  quantity  is  found  without  a  sign,  it  is  understood  to  be 
positive,  or  have  the  sign  +  prefixed. 

12.  Negative  Quantities,  are  those  which  are  to  be  subtrac- 
ted.    As  —  a,  or  —  2a6,  or  —  3a62 . 

13.  Like  Signs,  are  either  all  positive  (  -f-  ),  or  all  negative 

( -  )• 

14.  Unlike  Signs,  are  when  some  are  positive  (  +  ),  and 
ethers  negative  (  —  ). 

16.  The  Co-efficient  of  any  quantity,  as  shown  above,  is  the 
number  prefixed  to  it.     As  3,  in  the  quantity  3a6. 

16.  The  Power  of  a  quantity  (a),  is  its  square  (a^),  or 
cube  (a 3),  or  biquadrate  (a*),  &c.  ;  called  also,  the  2d  power, 
or  3d  power,  or  4th  power,  &c. 

17.  The  Index  or  Exponent,  is  the  number  which  denotes 
the  power  or  root  of  a  quantity.  So  2  is  the  exponent  of 
the  square  or  second  power  a^  ;  and  3  is  the  index  of  the 

cube  or  3d  power  ;  and  i  is  the  index  of  the  square  root,  a^ 

or  ^  a;  and  ^  is  the  index  of  the  cube  root,  a^,  or  ^  a. 

18.  A  Rational  Quantity,  is  that  which  has  no  radical  sign 
(  ^  )  or  index  annexed  to  it.     As  a,  or  3a6. 

19.  An  Irrational  Quantity,  or  Surd,  is  that  of  which  the 
value  cannot  be  accurately  expressed  in  numbers,  as  the  square 
roots  of  2,  3,  6.     Surds  are  commonly  expressed  by  means  of 

the  radical  sign  ^,  as  y'S,  ^  a,  ^a^ ,  or  ab^, 

20.  The  Reciprocal  of  any  quantity,  is  that  quantity  in- 
verted, or  unity  divided  by  it.  So,  the  reciprocal  of  a,  or 
a         1  a        b 

— ,  is  — ,  and  the  reciprocal  of  —  is  — . 
la  ha 

^  21.  The 


174  ALGEBRA. 

21.  The  letters  by  which  any  simple  quantity  is  expressed, 
may  be  ranged  according  to  any  order  at  pleasure.  So  the 
product  of  a  and  6,  may  l)e  either  expressed  by  a6,  or  ba ; 
and  the  product  of  a,  b,  and  c,  by  either  a6c,  or  acfc,  or  bac, 
or  bca,  or  cab.  or  c6a  ,  .as  it  matters  not  which  quantities  are 
placed  or  multiplied  first.  But  it  will  be  sometimes  found 
convenient  in  long  operations,  to  place  the  several  letters 
according  to  their  order  in  the  alphabet,  as  abc,  which  order 
also  occurs  most  easily  or  naturally  to  the  mind. 

22.  Likewise,  the  several  taembers,  or  terms,  of  which  a 
compound  quantity  is  compos^id,  may  be  disposed  in  any  or- 
der at  pleasure,  without  altering  the  value  of  the  signification 
of  the  whole.  Thus,  3a  —  tab  -}-  4fl6c  may  also  be  written 
3a  -f  4a6c  —  2a6,  or  Aabc  -f  3a  —  Saj^or  —  2a6  -f-  3a-|-4afec, 
&c.  ;  for  all  these  represent  the  sain6  thing,  namely,  the 
quantity  which  remains,  when  the  quantity  or  term  2a6  is  sub- 
tracted from  the  sum  of  the  terms  or  quantities  3a  aftd  4afec. 
But  it  is  most  usual  and  natural,  to  begin  with  a  positive  term, 
and  with  the  first  letters  of  the  alphabet. 

SOME  EXAMPLES  FOR  PRACTICE. 

fing  the    numeral   values   of  various  expressions,   or 
combinations,  of  quantities. 

Supposing  a  =  6,  and  6  =  6,  and  c  =a  4,  and  ci  =  1,  and 
«  =  0.     Then 

1.  Will  a2  4-  Sab  —  c^  —  36  +  90  —  16  =  llO. 

2.  And  2a3  —  3aH  -f  c^  =  432  —  640  -f  64  =  —  44. 

3.  And  a2   X  a+T—  2aic  =36  X  11  —  240  =  166. 

a3  216 

4.  And f-  c2  = {-16  =  12+16=  28. 

a -f  3c  18 


5.  And  ^2ac  -\- c^  or  2ac  +  c^  ^  "  =  ^  64  =  8. 

26c  40 

6.  And^cHh-;^^-^  =  2  +  -  =  7. 


.^h^—ac       36—1        35 
--/fea  -ra^'^12  — 7~  6 
8.  And  v'  fc2— ac4-^2flc  +  c2  =  1  +  8  =  9. 


9.  And  ^62— ac  +  ^2ac+c3  =^26  —  ^4  +  8  =  3. 

10.  Anda26  +  c  — d=  183. 

11.  And  9a6—  106^  +  c  =  24. 

*^  .      12.  And 


^ 


12. 

ADDITION, 
And X  d  =  45. 

c 

13. 

a+b      b 

And X-^-  13|. 

c         d 

14. 

a-\'b      a—b 

And =  If. 

c        •    d 

15. 

a^b 
And h  c  =  43. 

c 

16. 

And X  c  =  0. 

c 

175 


17.  And6— cX(^-e  =  1. 

18.  Anda-f6— c  -(^  =  8. 

19.  Anda-f6-.c-"tZ  =  6. 

20.  AndaStX-is  =  144. 

21.  And  ac«j(-r£  =  23. 

22.  Anda2c-ffe2e4.f^=:  i. 

23.  And X =  18^. 

d— c     c — d 


24.  And  y'a^b^  -^  a^  —6a  =  4-4936249. 

25.  And  3ac2  -{-  3/^3 -P"=  292-497942. 

26.  And  4a2—3a  ^/a^  —  fa6  =  72. 


ADDITION. 

Addition,  in  Algebra,  is  the  connecting  the  quantities 
together  by  their  proper  signs,  and  incorporating  or  uniting 
into  one  term  or  sum,  such  as  are  simiUr,  and  can  be  united. 
As,  3a  4-  26  -  2a  =  a  4-  26,  the  sum. 

The  rule  of  addition  in  algeJjra,  may  be  divided  into  three 
cases  :  one  when  the  quantities  are  like,  and  their  signs  like 
also  ;  a  second,  when  the  quantities  are  like,  but  their  signs 
unlike  ;  and  the  third,  when  the  quantities  are  unlike. 
Which  are  performed  as  follows. 

CASE 


•  *  The  reasons  on  which  these  operations  are  founded,   will  rea- 
dily appear,  by  a  little  reflection  on  the  nature  of  the  quantities  to 


176  ALGEBRA. 

CASE  L 

When  the  Quantities  are  Like,  and  have  Like  Signs. 

Add  the  co-efficients  together,  and  set  down  the  sum  5 
after  which  set  the  common  letter  or  letters  of  the  like  quan- 
tities, and  prefix  the  common  sign  -f-  or  —  . 


be  added  or  collected  together.  For,  with  regard  to  the  first  ex- 
ample, where  the  quantities  are  3a  and  5a  whatever  a  represents 
in  the  one  term,  it  will  represent  the  same  thing,  in  the  other  ;  so 
that  3  times  any  thing  and  5  times  the  same  thing,  collected 
together,  must  needs  make  8  times  that  thing.  As  if  a  denote  a 
shilling  ;  then  3a  is  3  shillings  ;  and  5a  is  5  shillings*  and  their  sum 
8  shillings.  In  like  manner,  —  2ad  apd  —  7a6,  or  —  2  times  any 
thing,  and  —  7  times  the  same  thing,  make  —  9  times  that  thing. 

As  to  the  second  case,  in  which  the  quantities  are  like,  but  the 
signs  unlike  ;  the  reason  of  its  operation  will  easily  appear,  by  re- 
flecting, that  addition  means  only  the  uniting  of  quantities  together 
by  means  of  the  arithmetical  operations  denoted  by  their  signs  4-  »nd 
— ,  or  of  addition  and  subtraction  ;  which  being  of  contrary  or  oppo- 
site natures,  the  one  co-efficient  must  be  subtracted  from  the  other,  to 
obtain  the  incorporated  or  united  mass. 

As  to  the  third  case,  where  the  quantities  are  unlike,  it  is  plain 
that  such  quantities  cannot  be  united  into  one,  or  otherwise  added, 
than  by  means  of  their  signs  ;  thus,  for  example,  if  a  be  supposed 
to  represent  a  crown,  and  6  a  shilling  ;  then  the  sum  of  a  and  b 
can  be  neither  2a  nor  2b,  that  is  neithei-  2  crowns  nor  2  shillings, 
hut  only  1  crown  plus  1  shilling,  that  is  a  +  6. 

In  this  rule,  the  word  addition  is  not  very  properly  used  ;  being  much 
too  limited  to  express  the  operation  here  performed.  The  business  of 
this  operation  is  to  incorporate  into  one  mass,  or  algebraic  expression, 
different  algebraic  quantities,  as  far  as  an  actual  incorporation  or 
union  is  possible  ;  and  to  retain  the  algebraic  marks  for  doing  it,  in 
eases  where  the  former  is  not  possible.  When  we  have  several  quanti- 
ties, some  affirmative  and  some  negati^  e  ;  and  the  relation  of  these 
quantities  can  in  th€  whole  or  in  part  be  discovered  ;  such  incorpora- 
tion of  two  or  more  quantities  into  one,  is  plainly  efiected  by  the  fore- 
going rules. 

It  may  seem  a  paradox,  that  what  is  called  addition  in  algebra, 
should  sometimes  mean  addition,  and  sometimes  subtraction.  But 
the  paradox  wholly  arises  from  the  scantiness  of  the  name  given  to 
the  algebraic  process  ;  from  eihploying  an  old  term  in  a  new  and  more 
enlarged  sense.  Instead  of  addition,  call  it, incorporation,  or  union, 
or  striking  a  balance,  or  any  name  to  which  a  mote  extensive  idea  may 
be  annexed,  than  that  which  is  usually  implied  by  the  word  addition  ; 
and  the  paradox  vanishes. 

Thus, 


ADDITION.  177 

Thus,  3a  added  to  5a,  makes  8  a. 

And  —  2ab  added  to  —  7a6,  makes  —  9ab. 

And  5a  -{-  76  added  to  7a  +  36,  makes  I2a  -{-  106. 

OTHER  EXAMPLES  FOR  PRACTICE. 


I 


3a 

—  36x 

bxy 

9a 

—  56a; 

2bxy 

5a 

—  46a; 

bbxy 

12a 

—  262; 

bxy 

a 

—  76a; 

3bxy 

2a 

—    bx 

6bxy 

32a 

—226a; 

ISbxy 

3z 

3^2  ^  sxy 

2ax  —  4y 

2z 

j?2  4.     ry 

4ax  —    y 

4z 

2x^  -f  4xy 

ax  —  3y 

z 

5x2  ^  2xy 

5ax  —  5y 

hz 

4x'i  +  3xy 

lax  —  2y 

\bz 

15a:3  -f  15^2/ 

19aj? —  15y 

5xy 

—   12?/2 

4a  — 46 

Uxy 

-    V 

5a  — 56 

22xy 

-    22/^ 

6a—    6 

llxy 

-    4y2 

3a  — 26 

H^y 

-   r 

2a  — 76 

i^y 

-    3y2 

8a—    6 

30  — 

13a;i  —  3:1-^ 

5xy 

—  3:r  -f-  4a6 

28  — 

lOxi  —  4xy 

Sxy 

—  4x-\-  3ab 

14  — 

Ux^  —  7^:?/ 

3xy 

—  5x  -f-  5a6 

10  — 

16a;^  —  5xy 

xy 

—  2x  +     ab 

16  — 

20a;i—    j:^/ 

Axy 

—    a;  4-  7a6 

Vot.  I.  2^  CASE 


17B  ALGEBRA. 

CASE  11. 

fVfien  the  Quantities  are  Likcy  but  have  Unlike  Signs ; 

Add  alf  the  affirmative  co-efficients  into  one  suno,  and  all 
rtie  negative  ones  into  another,  when  there  are  several  of  a 
kind.  Then  subtract  the  less  sum,  or  the  less  co-efficient, 
from  the  greater,  and  to  the  remainder  prefix  the  sign  of  the 
greater,  and  subjoin  the  common  quantity  or  letters. 

So     -|-  5a,  and  —  3a,  united,  make  -{-  2a. 
And  —  5a,  and  +  3a,  united,  make  —  2a, 

OTHER  EXAMPLES  FOR  PRACTICE. 


—  5a 
+  4a 
+  6a 
—.3a 
+    a 

+  Saxs 
4-  4aa:2 

—  8ax2 

—  6ax2 
-i-  5aa;2 

+     8a;3  4-  3y 

—  6a;»  4  4y 

—  16a;3  4-  5^ 
4-    3x^  —  ly 
+    2a;3  —  2y 

+  3a 

—  2aa;2 

—    8x3  _{.  sy 

— .    3a2 

—  5a2 

—  10a2 
+  10a2 
+  14a2 

H-      3621/3 

4-    9622,3 

— .    1062^3 

—  1962i/s 

—  2621/3 

-f  4a6  4-     4 
—  4a6  -\-  12 
4-  7a6  —  14 
4-    a6  -f-    3 
_  6a6  —  10 

^  3a:«:2 
4-     ax± 
+  bax.^ 
—  6axi 

4-   10^  ax 

—  3^ax 
4-     4^a:c 

—  12^aa: 

-1-   3?/  +  4ax^ 

—  2/ "—  ^(^^\ 
+   4y'-\'  2ax^ 

—  22/4-  6aar| 

CASE 


ADDITION.  119 

CASE  UI. 

Whenjhe  Quantities  are  Unlike, 

Having  collected  together  all  the  like  quantities,  as  in  the 
two  foregoing  cases,  set  down  those  that  are  unlike,  one  after 
another,  with  their  proper  signs. 

EXAMPLES. 

I 

3xy  Sjry— 12^2  4a:r  —  1 30  +  Sx^ 

2aj?  —  4^2  4-   3xy  >5x^  -f  Sax  +  9x^ 

— 5xy  +  4:r2  —  2xy  Ixy  —  Ax^  -{-  90 

6ax  —  3xy  -f-  4ar3  ^a:  +  40   —  6a:« 


—  ^xy  +  8a J7  Axy  —  ^x^  lax  +  \St^  •\-lry 


9  '\-  \0^  ax —  6y 
2r  4-  1  y/xy  -f  5y 
%  H-  3  ^  ax  —  4y 
10  —    4  ^  ax  +  4y 


9x^y 

14aa; 

—  2a;2 

-^Ix^y 

Saa;  •{-  3xy 

+  Saxy 

8y3 

—  4aa; 

—4x2  2/ 

3a:2 

+  26 

4a;2y 

4y  :r 

-    3y 

—  6xy2 

2^ary 

-{-  14a: 

+  3^2^ 

3x     + 

22/ 

-7a:2y 

—  9  +  2^.Ty 

3a2  +     9    -f    X2-.4 
2a    —    8    -I-  2a2  —  Sx 
4x^  —  2a2  4-  18    —  7 
-12    4-    a    —  3x2— 2y 


Add  a  +  6  and  3a  —  56  together. 
Add  6a  —  8x  and  3a  —  4x  together. 
Add  6a:  -66  +  a  +  8  to  -  5a  —  4jc  +  46  -  3. 
Add  a  4-  26  —  3c  -  10  to  36  —  4a  H-  5c  4-  10  and  56  —  c. 
Add  a  4-  6  and  a  -  6  together. 

Add  3a  +  6-  10  to  c  —  rf  -  a  and  —  4c  -I-  ^a  —  3Z>  -  7. 
Add  3a2  -f  62  —  c  to  2a6  -  Sa^  -f-  6c  -  6. 
Add  a3  -I-  62c  —  62  to  a62  —  a6c  -f  62. 

Add  9a  -  86  +  lOx  -  6<^  -  7c  -f  60  to  2x  -  3a  -  5c  -{-  4fc 
4-  6d  -  10. 


BUBTRAC- 


180 


ALGEBRA. 


SUBTRACTION. 

Set  down  in  one  line  the  first  quantities  from  which  the 
subtraction  is  to  be  made  ;  .and  underneath  them  place  all  the 
other  quantities  composing  the  subtrahend  :  ranging  the  like 
quantities  under  each  other,  as  in  Addition. 

Then  change  all  the  signs  (-f-  and  — )  of  the  lower  line,  or 
conceive  them  to  be  changed  ;  alter  which,  collect  all  the 
terms  together  as  in.the  cases  of  Addition*. 


From     7a-  —  36 
Take     3a^  —  86 


EXAMPLES. 


6x^  -f  5?/  --•  4 


80:^  —  3  +  60;—  y         j 
4xy  —  7  —  6a;  — 4y 


Rem.     4a2  -f  56  Sx^  —  61/+ 12         4xy  +  4-{-na:  -{-3y 


From     5xy  —   6 
Take— 2x2/  +    ^ 

Rem.     '7xy~\2  2y^  ^  by  —  S      -^2S-i'3x—8xy-{'2ay 


4y2^Sy-^4  — 20  — 6a;— So;?; 

22/2  4-  2?/  -I-  4    '     oxy —  9x -\- S  ^2ay 


From     Sx^y-^G     5^xy -}- 2x^xy      7a;2-f2^  or— 18+36 
Take—  2x^y  +  2     l^xy  -\-  3-^^xy      9x2—12     +  66+0:2 

Rem. 


*  This  rule  is  founded  on  the  consideration,  that  addition  and  sub- 
traction are  opposite  to  each  other  in  their  nature  and  operation,  as 
are  the  signs  +  and  — ,  by  which  they  are  expressed  and  represent- 
ed. So  that,  since  to  unite  a  negative  quantity  with  a  positive 
one  of  the  same  kind,  has  the  effect  of  diminishing  it,  or  subducting 
an  equal  positive  one  from  it,  tlierefore  to  subtract  a  positive  (which 
is  the  opposite  of  uniting  or  adding)  is  to  add  the  equal  negative 
quantity.  In  like  manner,  to  subtract  a  negative  quantity,  is  the  same 
in  effect  as  to  add  or  unite  an  equal  positive  one.  So  that,  by  chang- 
ing the  sign  of  a  quantity  from  -f  to  — ,  or  from  —  to  -J- ,  changes 
its  nature  from  a  subdiictive  quantity  to  an  additive  one  ;  and  any 
quantity  is  in  effect  subtracted,  by  barely  changing  its  sign. 


Bxv 


MULTIPLICATION.  181 

bxy  -  30         7x3  ^  2(a-\'b)         3xy^   -}"  20  a^  (xy+10) 
Ixy  —  60         2x^  -  4  (a  4-  i)  ^a;^?/^  -f  \2a^  (xy+10) 


From  a  -f-  ^»  take  a— 6. 

From  4a  +  46,  take  b  -{■  a. 

From  4a  —  4b,  take  3a  +  56. 

From  8a  —  12x,  take  4a  —  3^:. 

From  2x  —  4a— 26  -f  5  take  8  -  66  -f  a  +  6j:. 

From  3a  4-  6  4-  c  —  J  —  10,  take  c  -{-  2a  —  rf. 

From  3a  -f  6  -I-  c  —  d  —  10,  take  6  —  19  +  3a. 

From  2a6-f-62  -.4c  +  6c-  6,  take  3a2  — .  c  +  6^. 

From  a3  -f  363c  -f  a63  — a6c,  take  6^  -f-  06^  —  a6c. 

From  nx  +  6a— 46  +  40,  take  46  —  3a  +  4r  -f  6rf— 10. 

From  2ar— 3a  -f  46  -f  6c-50,take  9a  -f-  x  +  66-6c— 40. 

From  6a— 46— 12c  +  12x,  take  2a;  —  8a  +  46- 5c. 


MULTIPLICATION. 

This  consists  of  several  cases,  according  as  the  factors  are 
simple  or  compound  quantities. 

CASE  1.    When  both  the  Factors  are  Simple  Quantities  : 

First  multiply  the  co-efficients  of  the  two  terms  together, 
then  to  the  product  annex  all  the  letters  in  those  terms,  which 
will  give  the  whole  product  required. 

Note*.  Like  signs,  in  the  factors,  produce  +  and  unlike 
signs — ,  in  the  products. 


EXAMPLES. 


♦  That  this  rule  for  the  signs  is  true,  may  be  thus  shown, 
1.  When  +  a  is  to  be  moltiplied  by  +  c  /  the  meaning  is,  that^-  a 
is  to  be  taken  as  many  times  asthere  are  units  in  c;  and  since  the  sum 
of  any  number  of  positive  terms  is  positive,  it  follows  that  +  a  x  +c 
makes  +  ac 

2.  When 


3  Oct 
2b 

ALGEBRA 

EXAMPLES. 

—3a     , 
4-26 

7« 
—  4c 

—  4a 

20ab 

^6ab 

~-28ac 

-f  24aa7 

4ac 

Ax 

—  ^x^y 
6x^y^ 

•^4xy 

—  l2aHc 

'       36fl2x2                            — 

+  4x^y^ 

4x 

—  ax 

+  3xy 
—  4 

—  dxyz 

—  5ajr 

CASE  n. 

When  one  of  the  Factors  is  a  Compound  Quantity. 

Multiply  every  term  of  the  multiplicand,  or  compound 
^quantity,  separately,  by  the  multiplier,  as  in  the  former 
case  ;  placing  the  products  one  after  another,  with  the 
proper  signs  ;  and  the  result  will  be  the  whole  product  re- 
quired. 


2-  When  two  quantities  are  to  be  multiplied  together,  the  result  will 
be  exactly  the  same,  in  whatever  order  they  are  placed  ;  for  a  times 
tT  13  the  same  as  c  times  a,  and  therefore,  when  — a  is  to  be  multiplied 
by  -f"  <^»  or  -|-  c  bv— .a  ;  this  is  the  same  thing  as  taking  — a  as  many 
times  as  there  are  units  in  4.  c ;  and  as  the  sum  of  any  number  of 
negative  terms  is  negative,  it  follows  that  «-.  a  X  +c,  or  -f.*^  X  •- 
c  make  or  produce—  ac.  * 

3.  When— a  is  to  be  multiplied  by  —  c :  here  —a  is  to  be  subtract- 
ed as  often  as  there  are  units  in  c  .•  but  subtracting  negatives  is 
the  same  thing  as  adding  affirmatives,  by  the  demonstration  of  the 
rule  for  subtraction  ;  consequently  the  product  is  c  times  a,  or  -|-  ac. 

Otherwise.  Since  a — a  =  0,  therefore  (a  — a)X  — ■  c  is  also  =  0, 
because  0,  multiplied  by  any  quantity,  is  still  but  0  ;  and  since  the  first 
term  of  the  product,  or  a  X  -•  c  is  ■=  —  ac,  by  the  second  case  j 
therefore  the  last  term  of  the  product,  01  —  a  x  —  c,  must  be-4-ac, 
to  make  the  sum  =ts  o»  or  —  ac  4-  ac=^0  ;  that  is,  —  a  x  -  c  «= 
+  ««• 

EXAMPLES. 


MULTIPLICATION'.  im 


EXAMPLES. 


5a  — Sc 
2a 

3ac  —  46 
3a 

2a2  —  3c  +  5 
6c 

10a3  —  6ac 

9a3c  ^  \2ab 

2a2  6c  — 36c4-56c 

12x  —  2ac 
4a 

25c  —  76 

—  2a 

4x  —  6  +  3a6 
2a6 

3c3  4-^ 

4xy 

10x2—3^2 
— ■4x2 

3a3— 2x2  — 66 

2(1X3 

CASE  nr. 

When  both  the  Factors  are  Compound  Quantities  ; 

Multiply  every  term  of  the  multiplier  by  every  term  of 
the  multiplicand,  separately  ;  setting  down  the  products  one 
after  or  under  another,  with  their  proper  signs  ;  and  add  the 
several  lines  of  products  all  together  for  the  whole  product 
required. 

a  -f  6  3x  4-  2y  2a:2  4-  xy—  2y^ 

«  4-  6  4x  —  5y  3x  —  3y 


«a  +  a6 

+  ab  +  b^ 

12x3  ^  Qjry 

-Ux2/— 10i/2 

12:r3-  Ixy-^  \0y^ 
x^  +y 

x*  4"  y^^ 

6x3  43a-3  2/-6x2/3 

-6a;2  2/-3a;2/3  46?/3 

«2  4-  2a6  4  b^ 

6x  3  -3x2  y  -  exy^  +6y^ 

a-{-6 
«  — 6 

a3  4a6+63 
a  —6 

a2  4-a6 
-a6-63 

a3  4-«^6+a63 
-a26-a62  —  6- 

a2    *    —63 

flS          *         *         _   ^r. 

J^nte. 


T84  ALGEBRA. 

Note,  In  the  multiplication  of  compound  quantities,  it  i» 
the  best  way  to  set  them  down  in  order,  according  to  the 
powers  and  the  letters  of  the  alphabet.  And  in  multiplying 
them,  begin  at  the  left-hand  side,  and  multiply  from  the  left 
hand  towards  the  right,  in  the  manner  that  we  write,  which  is 
contrary  to  the  way  of  multiplying  numbers.  But  in  setting 
down  the  several  products,  as  they  arise,  in  the  second  and 
following  lines,  range  them  under  the  like  terms  in  the  lines 
above,  when  there  are  such  like  quantities  ;  which  is  the 
easiest  way  for  adding  them  up  together. 

In  many  cases,  the  multiplication  of  compound  quantities 
is  only  to  be  performed  by  setting  them  down  one  after 
another,  each  within  or  under  a  vinculum  with  a  sign  of 
multiplication  between  them.  As'(a-f  6)  X  (a  —  t)  X  3aK 
or  a-f"^  .  0  —  6  .  3a6. 

EXAMPLES  FOR  PRACTICE.     ' 

1.  Multiply  lOac  by  2a.  Ans.  20a2C. 

2.  Multiply  3aa  —  26  by  36.  Ans.  ^a^-Qh^, 

3.  Multiply  3a  +  26  by  3a  —  26.  Ans.  9a2— 462. 

4.  Multiply  x^  —  xy  •\"y^  hy  X  -{-  y.  Ans.  x^  -f  y^. 

5.  Multiply  a3  -f  a^b\-ah^  4"  6^  by  a -6.     Ans.  a*-  64. 

6.  Multiply  a2  4.  a6  -f-  6^  by  a2  -  a6  +  62 . 

7.  Multiply  3a;2  ^2a;«/  +  5  by  x^  -\-  2a;i/— 6. 

*  8.  Multiply  3a2  -2aa;  -f  bx^  by  3a2  —  Aax  -  Ix^, 
9.  Multiply  3x3  4.  2^72^2  -f- 3^3  by  ^x^^dx^y'^  -J-  Sy^. 
10.  Multiply  a2  +  a6  -f  62  by  a-  26. 


division- 


Division  in  Algebra,  like  that  in  numbers,  is  the  converse 
of  multiplication  ;  and  it  is  performed  like  that  of  numbers 
also,  by  beginning  at  the  left-hand  side,  and  dividing  all  the 
parts  of  the  dividend  by  the  divisor,  when  they  can  be  so 
divided  ;  or  else  by  setting  them  down  like  a  fraction,  the 
dividend  over  the  divisor,  and  then  abbreviating  the  fraction 
as  much  as  can  be  done.  This  will  naturally  divide  into  the 
following  particular  cases. 

CASE 


DIVISION.  18i 

CASE  I. 

When  the  Divisor  and  Dividend  are  both  Simple  Quantitiei ; 

Set  the  terms  both  down  as  in  division  of  numbers,  either 
the  divisor  before  the  diridend,  or  below  it,  hke  the  deno- 
minator of  a  fraction.  Tbew  abbreviate  these  terms  as 
much  as  can  be  done,  by  canceUinjo;  or  striking  out  all  the 
letters  that  are  common  to  them  both,  and  also  dividing 
the  one  co-efficient  by  the  other,  or  abbreviating  them  after 
the  manner  of  a  fraction,  by  dividing  them  by  their  commoQ 
measure. 

Note.  Like  signs  jn  the  two  factors  make  4-  in  the  quo- 
tient ;  and  unlike  signs  make  —  ;  the  same  as  in  multipli- 
eation*. 

EXAMPLES.     , 

1.  To  divide  6a6  by  3a, 

Qah 

Here  Qah  -r-  3a  or  3a)  Qah  (  or =  2t. 

3a 

c  ahx  a 
2.  Also  c  -r-  c  =^^ —  =  1  ;  and  ahx  -r-  hxy  = =  — . 

c  bxy        y 

•3.  Divide  16^2  by  8j:.  Ans.  2x. 

4.  Divide  l^a-x^  by-*-  3a^x.  Ans.  —  4x. 

5.  Divide  —  1  bay^  by  3ay.  *  Ans.  —  by, 

9xy 

6.  Divide—  ISax^y  hy^Qaxz.  Ans. — '-, 

4z 


*  Because  the  divisor  multiplied  by  the  quotient,  must  produce  the 
dividend.     Therefore, 

1.  When  both  the  terms  are  -f-,  the  quotient  must  be  -f  ;  because 
+  in  the  divisor  X  4-  in  the  quotient,  produces  -|-  in  the  dividend. 

2.  When  the  terms  are  both  —  ,  the  quotient  is  also  +  ;  because 
—  in  the  divisor  X  +  in  the  quotient,  produces  —  in  the  dividend. 

3.  When  one  term  is  -f.  and  the  other  — ,  the  quotient  must  be  —  i 
because  -^  m  the  divisor  X  —  in  the  quotient  produces  —  in  the  divi- 
dend, or,-  in  the  divisor  x  +  in  the  quotient  gives —  in  the. dividend. 

So  that  the  rule  is  general ;  viz.  that  like  signs  give  -|- ,  and  unlike 
signs  give  — ,  in  the  quotient. 

Vol.  I.  2i 

CASE 


186  ALGEBRA. 

CASE  II. 

When  the  Dividend  is  a  Compound  Quantity,  and  the  Divisor 
a  Simple  one  : 

PirjDF  eypry  term  of  the  dividend  by  the  divisor,  as  in 
the  former  case. 

EXAMPLES. 

ab-^-b^        a-\-b 

1.  (afe+fca  )  -J-  2b,  or = =  ^a  +  ^b, 

2b  2» 

10a6  4-  ISao: 

2.  {lOab  -f  I5ax)  -j-  5o,  or —  =  26  +  3ar. 

6a 

3.  (30a2r  -  482:)  ^  z,  or =  30a  —  48. 

z 

4.  Divide  6a6  — 'Sajr  -f-  a  by  2a. 

5.  Divide  3x2  __  15  ^  g^  ij.  g^  |jy  3_j.^ 

6.  Divide  6a6c  -{-  12a6a?  —  90^6  by  3a6. 

7.  Divide  10a2x  —  16x2  _  25x  by  hx. 

8.  Divide  16a^ic  —  Xbacx^  -}-  5aa*  by  —  Sac. 

9.  Divide  15a  -f-  3ay  —  18y    by  21a. 
18.  Divide  —  20a6  -f  60a63  by  —  6a6. 

CASE  in. 

When  the  Divisor  and  Dividend  are  both  Compound  Quantities  . 

1.  Set  them  down  as  in  common  division  of  numbers, 
the  divisor  before  the  dividend,  with  a  small  curved  line 
between  them,  and  ranging  the  terms  according  to  the 
powers  of  some  one  of  the  letters  in  both,  the  higher 
powers  before  the  lower. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor,  as  in  the  first  case,  and  set  the  result  in  the 
quotient. 

3.  Multiply  the  whole  divisor  by  the  term  thus  found,  and 
subtract  the  result  from  the  dividend. 

4.  To  this  remainder  bring  down  as  many  terms  of  the 
dividesd  as  are  requisite  for  the  next  operation,  dividing  as 
befgre  ;  and  so  on  to  the  end,  as  in  common  arithmetic. 

JVoit. 


DIVISIO.V.  187 

Note.  If  the  divisor  be  not  exactly  contaiaed  in  the  divi- 
dend, the  quantity  which  remains  after  the  operation  is  fin- 
ished, may  be  placed  over  the  divisor,  like  a  vulgar  fraction, 
and  set  down  at  the  end  of  the  quotient  as  in  common  aritn-* 
metic. 


EXAMPLES. 


o3—  a6 


-a6  -f-  h^ 


-c)  a^  — 4a3c  4-  ^ac^  ^-c^  (a*  —Sac  -^  c^ 


-3a2c4  3oc2 


ac^  —  c' 


«— 2)  a»-6a3  -f-  12a— 8  (a^  -4a  -{-  4 
a3~2a3 


-4a3  4-  12a 
-4fl3  -{-    8a 


4a- 8 
4a-8 


ct  -|-  2-)  a3  4. 2r3  ^ci2  ^a^  -{-  ^2 
a3  4-  a22 


—  a^z  —  az2 


a2r2  ^2^3 
a2r2  4.^3 


^  +  ar) 


188  ALGEBRA. 

<» -f  x)  (X*  — 5x4  .(a3 — ($2 X -{- aa:^ —^  x^ -— 

o-i-x 
«4  4-  a^x 


—  asx 

—  3x4 

—  a^xs 

—  3x4 

+  ax3 

a2x2 

—  ax3  — 

—  ax3  — 

3x4 

X4 

— 

2x4 

EXAMPLES  FOR  PRACTICE. 

1.  Divide  a^  +  4ax  -f-  4x2  by  a  +  2ar.         Ans.  a  +  2x. 

2.  Divide  a^  —  3a2^  -|-  3az^  — z^  by  a  —  z. 

Ans.  a2_2ct2r4-  z^ . 

3.  Divide  1  by  1  +  a.  Ans.  I  —  a  +  a^  -^  a^  -\- kc. 

4.  Divide  12a:4  —  192  by  3a;  —  6. 

Ans.  4x3  -\-  8x^  +  16ar  -f  32. 

6.  Divide  a^  — da'^b  +  lOa^fea^iOasfes  -|-  5a64  —  6^  by 

a2  _  2a6  4-  ^2.  Ans.  a^  -_3a2  6  +  3a62  —63. 

6.  -Divide  48^3  __  960^2  _  64aaz  +  150a 3  by  2z  —  3a. 

7.  Divide  6^  — 364a-2  +362x4  ^^^  by  63  -  362a;+36a:2  — a;3, 

8.  Divide  a'^ ^x'^  by  a  —  x. 

9.  Divide  a^  -\-  ba^x  -\-  5a x"  -}-  x^  by  a  -^  x. 
.10.  Divide  a^  +  4^262  —  326^  by  a  +  26. 

11.  Divide  24a4  —  64  by  3a  —  26. 


ALGEBRAIC  FRACTIONS. 


Algebraic  Fractions  have  the  same  names  and  rules  of 
operation,  as  numeral  tractions  in  common  arithmetic;  as  ap- 
pears in  the  following  Rules  and  Cases. 

CASE 


FRACTIONS.  189 

CASE  L 

To  reduce  a  Mixed  Quantity  to  an  Improper  Fraction, 
Multiply  the   integer  by  the   denoniinator  of  the  fraction, 
and  to  the  product  add  the  numerator,  or  connect  it  with  its 
proper  sign,  -f-  or  —  ;  then  the  denominator  being  set  under 
this  sum,  will  give  the  improper  fraction  required. 
EXAMPLES. 
b 

1.  Reduce  3f  and  a  -  — to  improper  fractions. 

j: 
3X64-4       164-4       19 

First,  3f  = = =  —  the  Answer. 

5                  6            6 
b         a  X   or  "   b       ax  —  b 
And,  a  —  —  = = the  Answer. 

2.  Reduce  a  -\ and  a to  improper  fractions. 

b  a 

a2        a  X  6  +  a3       ab-^-a^ 

First,  a  -\ = = the  Answer. 

b  b  b 

z^  — a2       a3  —  2-2  -f-  a2       2(x2  —2-2 

And,  a  —  = =    ^  the  Answer. 

a  a  a 

3.  Reduce  6^  to  an  improper  fraction.  Ans.  y* 

3a  X  —  3a 

4.  Reduce  1  —  — to  an  improper  fraction.       Ans. 

X  X 

3ax  +  a2 

6,  Reduce  2a to  an  improper  fraction. 

4x 
4j:— 18 

6.  Reduce  12  -| to  an  improper  fraction. 

5x 
1  -  3a  -  c 

7.  Reduce  x  -\ to  an  improper  fraction. 

c 
2x3  —  3^ 

8.  Reduce  4  -f  2a: to  an  improper  fraction. 

5a 
CASE  U. 
To  Reduce  an  Improper  Fraction  to  a  •whole  or  Mixed  Quantity. 
Divide  the  numerator  by  the  denominator,  for  the  integral 
part ;  and  set  the  remainder,  if  any,  over  the  denominator, 
for  the  fractional  part  ;  the  two  joined  together  will  be  the 
mixed  quantity  required. 

EXAMPLES 


190  ALGEBRA. 

EXAMPLES. 

16         a6  +  a2 

1.  To  reduce  —  and to  mixed  quantities. 

3  b 

First,  L6  =  16  ^  3  =  6^,  the  Answer  required. 

ab  4-  0.^       a2 

And, =  a6-{-a*-r-6=a-{ .     Answer. 

b  b 

2ac-3a2  3aj:+4x2 

2.  To  reduce and to  mixed  quantities. 

c  a  -\-  a; 

2ac-3a2       3a2 

First, =  2a(?— 3a2  -j-  c  =  2a .    Answer. 

c  c 

3ax  +  4j:* or^ 

And, =  3ax  +  4x^  -i-  a-f  x  =  3a;  H .    Ans. 

a  -f-  ar  a+rc 

33         2ax--3x2 

3.  Reduce  —  and  • to  mixed  quantities. 

6  a 

3x2 

Ans.  6|,  and  2x  — . 

a 
4a^x         2a2  4-2&2 

4.  Reduce and' to  whole   or  mixed   quan- 

2a  a  —  b 

tities. 

3x^  —  32/2  2x3  —  22/3 

6.  Reduce ,  and to  whole  tr  mixed 

x  +  y  X  —y 

quantities. 

10a2_4a-f  6 

6.  Reduce  . to  a  mixed  quantity. 

ba 
15a3-f6a2 

7.  Reduce to  a  mixed  quantity. 

3a3  +  2a2  —2a -4 

CASE  III. 
To  Reduce  Fractions  to  a  Common  Denominator. 
Multiply  every  numerator,  separately,  by  all  the  denomi- 
nators except  its  own,  for  the  new   numerators  ;  and  all  the 
denomincitors  together,  for  the  common  denominator 

When  il>e  denominators  have  a  common  divisor,  it  will  be 
better,  instead  of  multiplying  by  the  whole  denominators,  to 
multiply  only  by  those  parts  which  arise  from  dividing  by  the 
common  divisor.  And  observing  also  the  several  rules  and 
directions  as  in  Fractions  iti  the  Arithmetic. 

EXAMPLES. 


FRACTIONS.  l&l 

EXAMPLES. 

a  b 

1.  Reduce  —  and  —  to  a  commoo  denominator. 

X  z 

a  h       az         hx 

Here  —  and  — =  —  and  — ,  by  multiplying  the  terms  of  the 

X  z       xz         xz 

first  fraction  by  z,  and  the  terms  of  the  2d  by  x. 
a     X  b 

2.  Reduce  — ,  —  and  —  to  a  common  denominator. 

X     b  c 

ax  b         abc     cx^  b^x 

Here  — ,  — ,  and  —  = ,  ,  and ,  by  muhiplying  the 

X     b  c         bcx     bcx  bcx 

terms  of  the  1st  fraction  by  6c,  of  the  2d  by  ex,  and  of  the 
3d  by  bx, 

2a         36 

3.  Reduce  —  and  —  to  a  common  denominator. 

X  2c 

4ac  3bx 
Ans. and . 

2ca:  2cx 
2a         3a4-26 

4.  Reduce  —  and  ■  to  a  common  denominator. 

6  2c  4ac         Sab  -f  26« 

Ans. and , 

26c  26c 

5a         36 

5.  Reduce  —  and  — ,  and  4d,  to  a  common  denominator. 

3x         2c 

lOac         9bx         24cdx 

Ans. and and . 

Sex  6cx  6cx 

5  3a  3a 

6.  Reduce  —  and  —  and  26  -j ,  to  fractions  having  a 

6  4  6 

206         18tt6         48624.72a 

common  denominator.         Ans. and and — _ 

246  246  .  246 

1  2a2         2a2  4-62 

7.  Reduce  —  and and to  a  common  denomi- 

3  4  a-i-6 

nator. 

3b        2c         d  . 

8.  Reduce  —  and  —  and  —  to  a  common  denominator. 

4aa        3a         2a 

CASE 


192  ALGEBRA. 

CASE  IV. 

To  find  the   Greatest   Common    Measure  of  the    Terms  of  a 
Fraction. 

Divide  the  greater  term  by  the  less,  and  the  last  divi^sor  by 
the  last  remainder,  and  so  on  till  nothing  remains  ;  then  the 
divisor  last  used  will  be  the  common  measure  required  ;  just 
the  same  as  in  common  numbers. 

But  note,  that  it  is  proper  to  range  the  quantities  according 
to  the  dimensions  of  some  letters,  as  is  shown  in  division. 
And  note  also,  that  all  the  letters  or  figures  which  are  com- 
mon to  each  term  of  the  divisors,  must  be  thrown  out  of 
them,  or  must  divide  them,  before  they  are  used  in  the 
•pe  ration. 

EXAMPLES. 

ab-^-b' 


1.  To  find  the  greatest  common  measure  of  - 


ac^  -f  6c2 
ab  -\-  b^)  ac^  -f-  bc^ 
OT  a  4-  b  )  ac2  4-  6c2  (c2 


Therefore  the  greatest  common  measure  is  a  +  ^» 

as  — ab^ 
2.  To  find  the  greatest  common  measure  of  - 


«a  -f2a6-f62)a3  -.  ab^{a 

a3  4-  2a^b  -^  ab^ 


^2a^b  — 2a63^  a^  +  2ab  -f  b^ 
er      «  +     6  )  a2  4-  2a6  -f  fc^  (a  4-  * 
a^  +    ab 


ab  -f  b' 
a6  +  b^ 

Therefore  a  -|-  ft  is  the  greatest  common  divisor. 

a2  —  4 
3.  To  find  the  greatest  common  divisor  of- 


ab+2b 

Ans.  a— 2. 


4.  Tt 


FRACTIONS.  m 


4.  To  find  the  greatest  common  divisor  of  - 


.6* 
Ans.  a2  —b^, 

5.  Find  the  greatest  com.  measure  of . 

6a5-{-lUa*x+5a*;c3 

CASE  V. 

To  Reduce  a  Fraction  to  its  Lowest  Terms, 

Find  the  greatest  common  measure,  as  in  the  last  pro- 
blem Then  divide  both  the  terms  of  the  fraction  by  the 
common  measure  thus  found,  and  it  will  reduce  it  to  its  lowest 
terms  at  once,  as  was  required.  Or  divide  the  terms  by  any 
quantity  which  it  may  appear  will  divide  them  both,  as  in 
arithmetical  fractions. 

EXAMPLES. 

ab  +  b^ 

1.  Reduce to  its  lowest  terms. 

ac2  -I-  6c2 

ab-i-b^)  «c2  -|-6c2 
or  a  -f  6  )  «c3  4-  6c3  (c* 


Here  ab  -f-  b^  is  divided  by  the  common  factor  b. 

Therefore  a  -f-  ^  is  the  greatest  common  measure,   am! 
ab  +  b^         b 

hence  a  -}-  b) =  — ,  is  the  fraction  required. 

ac^  -f  6c2       c2 
c3  —  62c 

2.  To  reduce : —  to  its  least  terms. 

c2-f-26c+63 
€2  ^2be'\-b^)c^  —  b^c{c 

c3 -f-26c2-{-63c 


-^^bc^^n^c)  c2  -f-  26c  -f  62 

or  c  4-  fe)  c2  -f-  26c  +  62  (c  -f  6 
02-1-    be 


6c +  62 
6c +  62 

Vol.,!.  2$  Therefore 


194  ALGEBRA. 

Therefore  c  -f  fc  is  the  greatest  common  measure,  and 
c3  — 62c  c^—bc 

hence  c  -f-  6) = is  the  fraction  required. 

C'-f  5^6c-f  6*         c  +  6 

c3— 63  c3  4-  6c-f  h" 

3.  Reduce to  its  lowest  terms.      Ans. . 

C4~62c2  c3  4-  6c2 

a2  — 62  1 

4.  Reduce to  its  lowest  terms.  Ans.  • 


a* —64  a«  4-6* 

a4--6* 

6.  Reduce to  its  lowest  terms. 

a3-3a264-iia6*— fc=* 
3a«  +  6a*c  +  3a^c^ 

6.  Reduce ■ — -  to  its  lowest  terms. 

a^c-^  3a2c«-|-3ac3-|-c* 
a3— «62 

7.  Reduce  — —  to  its  lowest  terms. 

o2  4-2a6  4-6« 

CASE  VI. 
To  add  Fractional  Q;uantities  together, 

Ip  the  fractions  have  a  common  denominator,  add  all  the 
numerators  together  ;  then  under  their  sum  set  the  common 
denominator,  and  it  is  done. 

If  they  have  not  a  common  denominator,  reduce  them  t& 
one,  and  then  add  them  as  before. 

EXAMPLES. 


1.  Let  —  and  —  be  given,  to  find  their  sum. 

3  4 

a        a       4a       Sa       la  )    - 

Here 1 = 1 =  —  is  the  sum  required. 

3         4       12       12       12 
ah  c 

2.  Given  — ,  — ,  and  — ,  to  find  their  sum. 

be  d 

a        b  c       acd       bbd        bcc       acd'{-bbd'{-bcc 

Here  — +  — +  — = + + = — 

bed        bed       bed       bed  bed 

the  sum  roauired. 

^  3.  Let 


FRACTIONS.  196 

*3 .  Let  «  —  —  and  b  H be  added  together: 

b  e 

3x2  Zaa:  Scx^  2a6jr 

Here  a h  *  H ==  « \- b  -\ — — 

b  c  be  be 

^bx—Scx^ 

'  =  a  -f-  fc  H the  sum  required. 

be 
4x         2x  206x  -f-  6fl^ 

4.  Add  —  and  —  together.  Ans. . 

3a         5b  16a6 

a    a  a 

5.  Add  — ,  —  and  —  together.  Ans.  f^a. 

3  4  6 

2a- 3         5a  9a^6 

6.  Add and  —  together.  Ans. . 

4  8  .8 

a4-3  2a— 5  14a-13 

7.  Add  2a  +  -: —  to  4a  H .      Ans.  Oa  +  -: . 

6  4  20      V 

3a2  a  4-  6 

8.  Add  Sttf  and and together. 

46            36 
6a          6a         3a  +  2 
'    Q,  Add  *- ,  and  —  and together. 

4  6  7 
3a  a, 

10.  Add  2a,  and  —  and  3  -f-  -  together. 

8  6' 

3a  6a 

U.  Add  Ba-^ and  2a— —  together. 

4  8 

CASE  VII. 

*  To  Subtraet  one  Fractional  Quantiiif  from  another. 

Reduce  the  fractions  to  a  common  denominator,  as  in  addi- 
tion, if  they  have  not  a  common  denominator. 

Subtract  the  numerators  from  each  other,  and  under  their 
difference  set  the  common  denominator,  and  the  work  is  done. 


*  In  the  addition  of  mixed  quantities,  it  is  best  to  bring  the  frac- 
tional parts  only  to  a  coromon  denominator^  and  to  annex  their  sr.ni  to 
the  sum  of  the  integers*  w'th  the  proper  sign.  And  the  same  rule 
may  be  observed  for  mix§d  quantities  in  subtraction  ^Iso. 

EXAMPLES. 


i^  ALGEBKA. 

EXAMPLES. 

3g  4a 

1.  To  find  the  difference  of —  and  — . 
4  7 

3a     4a       21a       16a         5a 

Here = : —  =  —  is  the  difference  requireel. 

4        7         28         28  28 

2a— 6  3a--4fe 

:2.  To  find  the  difference  of and . 

4c  36 

2a— A     3a  — 46         6a6 -366     12ac— 166c 

Here = = 

4c  36  126c  126c 

6a6— .  366-  12ac  +  166c 

—-  is  the  difference  required. 

126c 

10a         4a 

3.  Required  the  difference  of and  — . 

9  7 

3a 

4.  Required  the  difference  of  6a  and  — .. 

4 

f  5a  2a 

5.  Required  the  difference  of —  and  ^^. 

4  3 

26  3a  -f  e 

6.  Subtract  —  from . 

c  6 

2a  -f  6  4a  +  8 

7.  Take from -. 

9  '        6 

a-^36  2a 

8.  Take  2a from  4a  -f  — 


CASE  VIII. 

To  Multiply  Fractional  Quantities  together. 

Multiply  the  numerators  together  for  a  new  numerator, 
and  the  denominators  for  ^  new  denominator.* 

*  1.  When  the  rumera'tor  of  one  fraction,  and  the  denominator  of 
th^  other,  can  be  divided  by  some  quantity,  which  is  common  to  both, 
the  quotients  may  be  used  instead  of  them. 

2.  When  a  fraction  is  to  be  multiplied  by  an  integer,  theproduct  is 
found  either  by  multiplying  the  numerator,  or  dividing  the  denomina- 
tor by  it ;  and  if  the  integer  be  the  same  with  the  denominator,  the 
nwmerator  may  be  taken  for  the  product. 

IPAMPLES^ 


FRACTIONS.  1^ 

EXAMPLES. 

a  ta 

1 .  Required  to  find  the  product  of  —  and  — . 

8  5 

a  X  2a       2a2        a^ 

Here = =  —  the  product  required. 

8  X  5  40        20 

a     3a         6a 

2.  Required  the  product  of — ,  — ,  and  — . 

3  4  7 
a  X  3a  X  6a       ISa^       So^ 

= = the  product  required. 

3X4X7  84  14 

2a         a  4-  6 

3.  Required  the  product  of  —  and . 

b  2a-{-c 

2a  X  (a-f-6)       2aa  -f  2a6 

Here = the  product  re(luired. 

b  X  (2a+c)        2ab  -f  be 

4a         Ga 

4.  Required  the  product  of  —  and  — . 

3           5c 
3a         4b^ 
6.  Required  the  product  of —  aad . 

4  3a 
3a         8ac         4ab 

6.  To  multiply  —  and and together. 

6  6  3c 

ab         3a2 

7.  Required  the  product  of  2a  +  —  and . 

2c  b 

2a2  —262  4a2  +  26- 

8.  Required  the  product  of and— . 

36c  a  -I-  6 

2a  +  1         2a  -  1 

9.  Required  the  product  of  3a,  and and . 

a  2a  -f  6 

X       x^  a        a^ 

10.  Multiply  a  -h by  a; {- 

2a     4a2  2x       Ax^ 

CASE  IX. 
To  Divide  one  Fractional  (Quantity  by  another. 
Divide  the  numerators  by  each  other,  and  the  denomina- 
tors by  each  other,  if  they  will  exactly  divide.     But,  if  not, 
then  invert  the  terms  of  the  divisor,  and  multiply  by  it  exact- 
ly as  in  muItipHcation,  EXAMPLES. 

*  I .  I  f  the  fractions  to  be  divided  have  a  coniaion  denominator,  take  the  numerator  of  the 
dividend  for  a  new  numerator,  and  the  numerator  of  the  divisor  iVir  the  new  denominator. 

2.  When  a  fraction  is  t ,  be  divided  hy  any  quantttr.  it  is  the  stme  thing  whether  the 
munemiot  be  diYnied  t»y  it,  or  the  deooiaiaator  multiplied  by  it.  3.  When 


198  ALGEBRA. 

EXAMPLES.  ' 

a         .3a. 

1.  Required  to  divide  —  by — . 

4        8 
a        3a       a        8         8a         2 

Here r  —  =  —  X  —  — =  —  the  quotient.  . 

4         8        4        3a       I2a        3 
3a      5c 

2.  Required  to  divide  —  by — . 

26      4d 
3a       6c       3a      4a       VZad      6ad 

Here  —  -j =  —  X  —  =  — ^ —  =: the  quotient, 

26       4d      26       5c       106c       56c 
2a +  b       3a +  26 

3.  To  divide by .     Here, 

3a— 26       4a  +  6 
2a +  6     4a -f- 6       8a2  -|- 6a6 -f  62 

X — : = the  quotient  required. 

3a— 26      3a+26  9a^  -  46^ 

3a2  2a 

4.  To  divide  — by — . 

a2  4-  65»       2a  -f-  26 
3a2  a-f6         3a2X(a  +  6)  3a 

Here, X  • 


a2-f-63         a  (a3-f63)Xa      a^  —  a6 -f  6^ 

is  the  quotient  required 

3a;      U 

5.  To  divide  —  by  — .  ^ 

4         12 

6.  To  divide by  3x, 

5 
3a;  -f-  1       4x 

7.  To  divide by — . 

9  3^ 

4x  X 

8.  To  divide ^ — by—. 

2x- 1  3 

4x       3a 

9.  To  divide  —  by —. 

6        66 
2a— 6       5ac 

10.  To  divide by . 

4cd  6d 

6a* -56*  6a3-f6ai 

11 .  ©ivide by . 

2a2  -4a6  +  26^         4a  —  46  INVOLU- 

3.  When  the  two  numerators,  or  t)ie  two  denominators,  can  be  divided  by  jome  commvn 
quantity,  let  that  be  doae,  ami  the  quotients  qsqd  uute^id  of  the  fractiosa  fir«t  prepoted. 


[    m   ] 

INVOLUTION. 

iNVOLtJTioN  is  the  raising  of  powers  from  any  proposed 
root  ;  such  as  findin/j;  the  square,  cube,  biquadrate,  &,c.  of  any 
given  quantity.     '1  he  n>ethod  is  as  follows  : 

*  Multiply  the  root  or  given  quantity  by  itself,  as  m^ny 
tiiues  a.*  there  are  units  in  the  index  less  one,  and  the  last  pro- 
duct will  be  the  power  required. — Or,  in  literals,  multiply 
the  index  of  the  root  by  the  index  of  the  power,  and  the  result 
will  be  the  power,  the  same  as  before. 

JVote.  When  the  sign  of  the  root  is  -f,  all  the  powers  of  it 
will  be  -f  ;  but  when  the  sign  is  — ,  all  the  even  powers  will  be 
-i-,  and  all  the  odd  powers  —  ;  as  is  evident  from  multrplication. 
EXAMPLES. 


a,  the  root 

a2,  the  root 

a2  =  square* 

a*     =  square 

a3  =  cube 

a6    =  )cnhe 

a*  =  4th  power 

a  a    =  4th  power 

a^  =  5th  power 

ftio  =  5th  power 

&c. 

&c. 

—    2a,  the  root 

—     3a63 ,  the  root 

-|-     4a2  =  square 

-|-     9a2  6*  =  square 

—    8a  3  =  cube 

—  27a3  66  =  cube 

-f  16a*  =4th  power 

"--f  Sla46«  =  4th  power      ■ 

—  32a5  =  5th  power 

— 243as6»o=  5th  power 

2ax^ 

a 

,  the  root 

— ,  the  root 

36 

26 

4a3a;* 

a2 

-| =  square 

—  =  Square 

96^ 

463^ 

Sa^x^ 

a3 

—  cube 

—  ==  cube 

863 

2763 

16a4x» 

,     fl4 

-\ —  =  4th  power. 

816^ 

166* 

♦  A  ay  power  of  the  product  of  two  or  moie  quantities,  is  equal  t© 
the  same  power  of  each  of  the  factors,  multiplied  together. 

And  any  power  of  a  fi-action,  is  equal  to  the  san^e  power  of  the  nu- 
me  ator,  divided  by  the  I  ke  power  of  the  denominator. 

Also,  powers  or  roots  of  the  same  quantity,  are  multiplied  by  one 
another,  by  adding  their  exponents  j  or  divided,  by  subtracting  their 
exponents. 

a3  3-2 

Thus,a3  X  «2  =  a8*2  ^  a5.    And  a3 -r  a^  or  -  =  a      =«. 


200  ALGEBRA. 

X  —  a  =  root  ^  -|-  o  =  root 

X  —  a  X  -\-  a 


X*  —ax  x^-\-  ax 

—ax  -[-  a2  -\'  ax  '\-  a^ 


x^^   2ax  +  a2  square         x^-\-  2ax  -{-  a^ 
X  —  a  JT  -j-  a 


:r3—  3ax2  '{-Sa'^x—a^         x^-}-  3ax^ -{-Sa^x-^-a^ 


the  cubes,  or  third  powers,  of  x  —  a  and  x  -{-  a. 
EXAMPLES   FOR  PRACTICE. 

1.  Required  the  cube  or  3d  power  of  3a^. 

2.  Required  the  4th  power  of  2a^b. 

3.  Required  the  3d  power  of  — ■  4a^b^. 

a^x 

4.  To  find  the  biquadrate  of . 

263       ' 
6.  Required  the  6th  power  of  a  —  2x, 

6.  To  find  the  6th  power  of  2a2 . 

Sir  Isaac   Newton's    Rule  for   raising  a  Binomial  to  any 
Power  whatever*. 

1.  To  find  the  terms  mthout  the  Co-efficients.  The  index  of 
the  first,  or  leading  quantity,  begins  with  the  index  of  the 
given  power,  and  in  the  succeeding  terms  decreases  conti- 
nually by  1,  in  every  term  to  the  last ;  and  in  the  2d  or 
following  quantity,  the  indices  of  the  terms  are  0,  1,2,  3,  4, 
&c.  increasing  always  by  1 .  That  is,  the  first  term  will  con- 
tain only  the  first  part  of  the  root  with  the  same  index,  or  of 

*  This  rule,  expressed  in  general  terms,  is  as  follows  ; 

fifl\x)^  —  a"  +  n  .  a^-^x  +  «.— — ao~2x2-^M.—  .  —  an-^x^  &c. 
2  2  3 

n— 1  n— 1     n  — 2 

^a— x)»  «■  a"  -  n  .  an-lx+  n . — -^a^-^^x^  -  n  —  .  -^  a^-^x^  &c. 
2  2  3 

J^Tote.  The  sum  of  the  co-efficients,  in  every  power,  is  equal  to  the 
number  2,  when  raised  to  that  power.  Thus  1  +  1  =  2  in  the  first 
power  ;  1  +  2  +  1  =  4  -=  22  in  the  sljaare  ;l+3+3  +  l«'8=s 
23  in  the«ube,  or  third  power  ;  and  s«  on* 

the 


INVOLUTION.  201 

the  same  height  as  the  intended  power  :  and  the  last  term  of 
the  series  will  contain  only  the  2d  part  of  the  given  root, 
when  raised  also  to  the  same  height  of  the  intended  power  : 
hut  all  the  other  or  intermediate  terms  will  contain  the  pro- 
ducts of  some  powers  of  both  the  members  of  the  root,  in 
such  sort,  that  the  powers  or  indices  of  the  1st  or  leading 
member  will  always  decrease  by  1,  while  those  of  the  2d 
member  always  increase  by  1. 

2.  To  find  the  Co-efficients.  The  first  co- efficient  is  always 
1 ,  and  the  second  is  the  same  as  the  index  of  the  intended 
power  ;  to  find  the  3d  co-efficient,  multiply  that  of  the  2d 
term  by  the  index  of  the  leading  letter  in  the  same  term,  and 
divide  the  product  by  2  ;  and  so  on,  that  is,  multiply  the  co- 
efficient of  the  term  last  found  by  the  index  of  the  leading 
quantity  in  that  term,  and  divide  the  product  by  the  numl)er 
of  terms  to  that  place,  and  it  will  give  the  co-efficient  of  the 
term  next  following ;  which  rule  will  find  all  the  co-efficients, 
one  after  another. 

JVote.  The  whole  number  of  terms  will  be  1  more  than  the 
index  of  the  given  power  :  and  when  both  terms  of  the  root 
are  +,  all  the  terms  of  the  power  will  be  -{-  ;  /but  if  the  se- 
cond term  be  — ,  all  the  odd  terms  will  be  -4-,  and  all  the 
even  terms  — ,  which  causes  the  terms  to  be  4*  and  —  alter- 
nately. Also  the  sum  of  the  two  indices,  in  each  term,  is 
always  the  same  number,  viz.  the  index  of  the  required 
power  :  and  counting  from  the  middle  of  the  series,  both 
ways,  or  towards  the  right  and  left,  the  indices  of  the  two 
terms  are  the  same  figures  at  equal  distances,  but  mutually 
changed  places.  Moreover,  the  co-efficients  are  the  same 
numbers  at  equal  distances  from  the  middle  of  the  series, 
towards  the  right  and  left  ;  so  by  whatever  numbers  they 
increase  to  the  middle,  by  the  same  in  the  reverse  order  they 
decrease  to  the  end. 

EXAMPLES. 

1.  Let  a  -{-  jc  he  involved  to  the  5th  power. 

The  terms  without  the  co-efficients,  by  the    1st  rule, 
will  be 

and  the  co -efficients,  by  the  2d  rule,  will  be 

..5X4    10X3      10x25x1 
1,  6,  -y-,  — ^ — ,  — - — ,  -J-  ; 

or  1,  5,  10,         10,  5,  1  ; 

Therefore  the  5th  power  altogether  is 
a5  4-  5a*x  -I-  lOa^jps  ^  lOa^x^  -f-  dax*  -f  ««. 
Vol.  I.  27  But, 


202  ALGEBRA. 

But  it  is  best  to  set  down  both  tbe  co-efficients  and  the 
powers  of  the  letters  at  once,  in  one  line,  without  the  inter- 
mediate lines  in  the  above  example,  as  in  the  example  here 
below. 

2.  Let  o  —  X  be  involved  to  the  6th  power. 

The  terms  with  the  co-efficients  will  be 
a^  ~  ea^x  -f-  15a4:r2  —  20a^x^  +  Iba'^x'^  -  6aa;«  -f  x*. 

3.  Required  the  4th  power  of  a  —  x. 

Ans.  a*  —  4a^x  -f  Ga^x^  —  4ax^  -f  ^*. 
And  thus  any  other  powers  may  be  set  down  at  once,  is 
the  same  manner  ;  which  is  the  best  way. 


EVOLUTION. 


Evolution  is  the  reverse  of  Involution,  being  the  method 
of  finding  the  square  root,  cube  i^oot,  &c.  of  any  given 
quantity  whether  simple  or  compound. 

CASE  L 

To  find  the  Roots  of  Simple  Quantities, 

Extract  the  root  of  the  co-efficient  for  the  numeral 
part  ;  and  divide  the  index  of  tl.e  letter  or  letters,  by  the 
index  of  the  power,  and  it  will  give  the  root  of  the  literal 
part  ;  then  annex  this  to  the  former,  for  the  whole  root 
sought*. 


*  Any  even  root  of  an  affirmative  qiiantity,  may  be  either  +  or  —  : 
thus  the  square  root  of  +  as  is  either  -f-  a,  or  —  as  because  -j-  aX 
-f.  a  =  -f.  a,  and  —  aX  —  a  =  -^  ais  also. 

But  an  odd  root  of  any  quantity  will  have  the  same  sign  as  the 
quantity  itself:  thus  the  cube  root  of  +  a3  is  +  a  and  the  cube 
i-cot  of  a3  is  -  a  ;  for  -f  a  X  +  a  X  -f*  a  =r  +  a3,  and  —  rt  X  — 
a  X  —  a  =  —  «3, 

Any  even  root  of  a  negative  quantity  is  impossible  ;  for  neither 
-f  a  X  +  fl,  nor  —  a  X  —  a  can  produce  —  flS. 

An .  root  of  a  product,  is  equal  to  the  like  root  of  each  of  the  fac- 
tors multiplied  togetlier.  And  for  the  root  of  a  fraction,  take  tlie  root 
of  the  numerator,  and  the  root  of  the  denominator. 

EXAMPUIS. 


EVOLUTION.  20S 


EXAMPLES. 

1.  The  square  root  of  4a^ ,  is  2a. 

2.  The  cube  root  of  Sa^,  is  2a^  or  2a. 

Sii^b^  5j2b2    .     ab 

3.  The  square  root  of  -^,  or  v/"^^  >  ^^Yc  ^  ^' 

4.  The  cube  root  of ^n^^^' sT  ^ 

6.  To  find  the  square  root  of  ^a^bK  Ans.  ab^  ^2. 
^  6.  To  find  the  cube  root  of  —  64a^b^  Ans.  —  4ab^. 

7.  To  find  the  square  root  of  Arj-'  ^^^'  2a^  >/  -3^- 

8.  To  find  the  4th  root  of  Sla^ft^.  Ans.  Sab  ^  b. 

9.  To  find  the  6th  root  of  -  32a^b\  Ans--2a6  ^  ^• 

CASE  II. 

To  find  the  Square  Root  of  a  Compound  Quantity . 

This  is  performed  like  as  in  numbers,  thus  : 

1.  Range  the  quantities  according  to  the  dimensions  of 
one  of  the  letters,  and  set  the  root  of  the  first  term  in  the 
quotient. 

2.  Subtract  the  square  of  the  root,  thus  found,  from  the 
first  term,  and  brin^  down  the  next  two  terms  to  the  re- 
mainder for  a  dividend ;  and  take  double  the  root  for  a 
divisor. 

3.  Divide  the  dividend  by  the  divisor,  and  annex  the  re- 
sult both  to  the  quotient  and  to  the  divisor. 

4.  IVIultiply  the  divisor  thus  increased,  by  the  term  last 
set  in  the  quotient,  and  subtract  the  product  from  the  divi- 
dend. 

And  so  on,  always  the  same,  as  in  common  arithmetic. 

EXAMPLES. 

p       1.  Extract  the  square  root  of  a4—4a3i-f-6a2  62— 4a63-f  6*. 
a*  — 4a36  -f  ^a^b^  —  4ab^  -f-  b*{a^  —  2ab  +  6^  the  root. 

2a3  —  2ab)  -  4a^b  -}-  6a^b^ 
-  4a^b  4-  4a362 


2a2  —  4a&-i-62)  2an^  -  4ab^ -i- b^ 
2a2  62.-4a63  4-64 


2.  Find 


204  ALGEBRA. 

2.  Find  the  root  of  a*  -f  4a^b  +  10a3  62  4-  i2ab^  -f  96". 
a*  -f  4a^b  +  10a2  62  +  12ab^  -f  96*  (aa+  2a6  -f  3fe2. 


2a2  +  2a6  )  4aH  +  lOaafe^ 
4a36  4-     4a2  62 


2a2  +  4a6-|-363  )  60262  +  1206  3  + 9^* 
6a2  62-f  12a63-|-96* 


3.  To  find  the  square  root  of  a*  +  4a^  +  6a^  -|-  4a  +  1. 

Ans.  a2  -|-2a  +1. 
4«  Extract  the  square  root  of  a*  —  2a^  +  2a2  —  a  +  a. 

Ans.  x^  —  X  -f"  i- 

5.  It  is  required  to  find  the  square  root  of  a^  —  ab. 

6       62  d3  „ 

Ans.  a  —  -  — &c. 

2        8a  Jba2 


To  find  the  Roots  of  any  Powers  in  General. 

This  is  ako  done  like  the  same  roots  in  numbers,  thus  ; 

Find  the  root  of  the  first  term,  and  set  it  in  the  quotient. 
—Subtract  its  power  from  that  term,  and  bring  down  the* 
second  term  for  a  dividend. — Involve  the  root,  last  found,  to 
the  next  lower  power,  and  multiply  it  by  the  index  of  the 
given  power,  for  a  divisor. —  l/ivide  the  dividend  by  the  di- 
visor, and  set  the  quotient  as  the  next  term  of  the  root. — 
Involve  now  the  whole  root  to  the  power  to  be  extracted  ; 
then  subtract  the  power  thus  arising  from  the  given  power, 
and  divide  the  first  term  of  the  remainder  by  the  divisor  first 
found  ;  and  so  on  till  the  whole  is  finished*. 

EXAMPLES. 


♦  As  this  method,  in  high  powers,  may  be  thought  too  laborious,  it 
■will  not  be  improper  to  observe,  that  the  roots  of  compound  quantities 
may  sometimes  be  easily  discovered,  thvis  : 

Extract  th-  roots  of  some  of  the  most  simple  terms,  and  connect 
them  together  by  the  sign  +  or  — ,  as  may  be  judged  most  suitable 
for  the  purpose Involve  the  compound  root,  thus  found,  to  the  pro- 
per power  i  then,  if  this  bf  the  same  wivh  the  given  quantity,  it  is  the 

root  required But  if  it  be  found  to  difrer  only  in  some  of  the  signs, 

change  them  from  -f  to  — ,  or  from  —  to  +,  till  its  power  agrees 
witii  the  given  one  throughout. 

,  Thus 


EV6LUT101I5J.  .  20b 

EXAMPLES. 


1.  To  find  the  square  root  of  a* -Sa^fc+Sasfcz— 2a6a+fe<. 
a*  —  2a^b  -f  Sa^fc^  —  Saft^  4-6*  (a3~  06  -f  b^ 


•  2a^b 


2.  Find  the  cube  root  of  a^  —  6a«  +  21a*  —  44a=»  +  63a« 

~54a  -f-  27. 
a6-6a5  4-21a*-44a3  +  63a3— 54a  +  27  (a3-.2a  4- 3. 
a6 


3a4  )  —  6a5 


a6_6a5  -f  12a4— 8a3  =  (a^  -.2a)» 

3a4  )  -j-  na* 
a6_6a:5-|-21a4— 44a34-63a2~54a  +  27  =  (a2— 2a— 3)^-. 


3.  To  find  the  square  root  of  a^  —  2ab  +  2aa;  +  62  — 
2bx  -\~  x^-  Ans.  a  —  b  -^  x, 

4.  Find  the  cube  root  of  a6-3a5  -{-  9a4  —  iSa^  -f  18a2  — 
12a  +  8.  Ans.  a2-.a  -f-  2. 

5.  Find  the  4th  root  of  81a*  — 216a36  -f-  216a2  62  _  geab'-' 
+  166*.  Ans.  3a  —  26. 

6.  Find  the  5th  root  of  a«  —  10a*  -{-  40a3  -  SOa^  -f  80ft 
—  32.  Ans.  a  —  2. 

7.  Required  the  square  root  of  1  —  x^ . 

8.  Required  the  cube  root  of  1  — x^. 


Thus,  in  the  5th  exninple,  the  root  3a  —  26,  is  the  difference  of 
the  roots  of  the  first  and  last  terms  ;  and  in  the  third  example,  the 
root  a  ~  6  4-  >r,  is  the  sOrri  of  the  roots  of  the  1st,  4th,  aiid  6lh  terms. 
The  same  may  ulso  be  observed  of  the  6th  example,  where  the  root  is 
foiyjd  from  the  first  and  last  terms. 

SURDS. 


2e6  ALGEBRA. 


SURDS. 


Surds  are  such  quantities  as  have  not  exact  values  in  num- 
bers ;  and  are  usually  expressed  hy  fractional  indices,  or  by 

means  of  the  radical  sign  y/.     Thus,  3^,  or  ^  3,  denotes  the 

square  root  of  3  ;  and  2^  or  »/  22,  or  ^  4,  the  cube  root  of 
the  square  of  2  ;  where  the  numerator  shows  the  power  to 
which  the  quantity  is  to  be  raised,  and  the  denominator  its  root, 

PROBLEM  I. 

To  Reduce  a  Rational  Quantity  to  the  Form  of  a  Surd, 

Raise  the  given  quantity  to  the  power  denoted  by  the  index 
•f  the  surd  ;  then  over  or  before  this  new  quantity  set  the 
radical  sign,  and  it  will  be  the  form  required. 

EXAMPLES, 

1.  To  reduce  4  to  the  form  of  the  square  root. 
First,  42  =  4  X  4  =  16  ;  then  ^  16  is  the  answer. 

2.  To  reduce  3a^  to  the  form  of  the  cube  root. 
First  3a2  X  3a2.  x  3a2  =  (3a2)3  =  27a«  j 

then  3/  27a6  or  (27a6)  3-  is  the  answer, 

3.  Reduce  6  to  the  form  of  the  cube  root. 

Ans.  (216)2  or  ^216, 

4.  Reduce  ^ab  to  the  form  of  the  square  root. 

Ans.  ^  ^a^b''^ 

5.  Reduce  2  to  the  form  of  the  4th  root.  Ans.  (16)*. 

6.  Reduce  a^  to  the  form  of  the  5th  root. 

7.  Reduce  a  -j-  a;  to  the  form  of  the  square  root. 

8.  Reduce  a  —  a;  to  the  form  of  the  cube  root. 

PROBLEM  IL 

To  Reduce  Quantities  to  a  Common  Index. 

1.  Repuce  the  indices  of  the  a;iv<^n  quantities  to  a  common 
denominator,  and  involve  each  of  them  to  the  power  denoted 
by  its  numerator  ;  then  1  set  over  the  common  denominator 
will  form  the  common  index.     Or, 

2.  Ijf 


SUMS.  207 

^  f .  If  tbe  comnion  index  be  gives,  divide  the  indices  of  the 
quantities  by  the  given  index,  and  the  quotients  will  be  the 
new  indices  for  those  quantities.  Then  over  the  said  quan- 
tities, with  their  new  indices,  set  the  given  index,  and  thejr 
^ill  make  the  equivalent  quantities  sought. 

EXAMPLES. 

1.  Reduce  3*  and  6*  to  a  common  index. 
Here  1  and  }  =  j\  and  j\. 

Therefore  3^"  and  5^^°  =  (3s)^^znA  (62)t'»=^  3^  and  v' 6« 

=  ^  243  and  ^  26. 

2.  Reduce  a^  and  b^  to  the  same  common  index  |. 
Here,  f  -j-  i  =  i  X  f  =  f  the  1st  index. 

and  ^  4-  i  =  i  X  f  =  I  the  2d  index. 

I  a.  L  2 

Therefore(a«)^  and  (63)  2^  or  ^^  a^  and  y  6^  are  the  quaa- 

tities. 

3.  Reduce  4^  and  5^  to  the  common  index  i. 

Ans.  2563  )^  and  25  . 

4.  Reduce  o^  and  x*  to  the  common  index  a. 

Ans.  (a3)«  and(x^)^. 

5.  Reduce  a^  and  x^  to  the  same  radical  sign. 

Ans.  -^  a*  and  y/  x^. 

6.  Reduce  (a  -|-  ^)^  *°<^  (a  — x)^  to  a  common  index. 

7.  Reduce  (a  -f-  6)^  and  (0—6/  to  a  common  index, 

PROBLEM  ra. 

To  Reduce  Surds  to  more  Simple  Terms^ 

^iND  out  the  greatest  power  contained  in,  or  to  divide  the 
given  surd  ;  take  its  root,  and  set  it  betore  the  quotient  or  the 
remaining  quantities,  with  the  proper  radical  sign  between 
them. 

EXAMPLES. 

1.  To  reduce  ^  32  to  simpler  terms* 

Here  y  32  =  y  16X2  =  y  16  Xy2  =  4  X^2  =  4^2^ 

2.  To  reduce  ^  320  to  simpler  terms. 

3/320=  ^64X5  =  ^64X  V'^  =  4X  \/ 5^4  1/5. 

3.  Reduce 


208  AL,GEBRA. 

3.  Reduce  -^  75  to  it3  simplest  terms.  Ans.  5  ^  S. 

4.  Reduce  ^  f  |  to  simpler  terms.  Ans.  fj  y/  33. 

5.  Reduce  ^  109  to  its  simplest  terms.  Ans.  ^  37. 

6.  Reduce  ^  *^^  to  its  simplest  terms.  Ans.  -|-  ^  10. 

7.  Reduce  y/  Iba^b  to  its  simplest  terms.       Ans.  ba  ^  36. 
jVyic,  There  are  other  cases  of  reducing  algebraic  surds 

to  simpler  forms,  that  are  practised  on  several  occie^ions  ;  one 
instance  of  which,  on  account  of  its  simplicity  and  nsetulness, 
may  be  here  noticed,  viz.  in  fractional  forms  having  com- 
pound surds  in  the  denominator,  multiply  both  numerator  and 
denominator  by  the  same  terms  of  tht  denominator,  hut  hav- 
ing one  sign  changed,  from  +  to— or  from  — to  -}■»  which  will 
reduce  the  fraction  to  a  rational  denominator. 

Ex.  To  reduce ,  multiply  it  by ,  ^nd 

V'ft  — v/3                            a/54--v/3 
16  -f  2  ^  15                                      Sy'lS- V5 
it  becomes ==  8  -f  ^  16.  Also,  if ; 

^15—^5  65  —  7^/76 

multiply  it  by ,   and  it  becomes = 

^\b^^b  15  —  5 

65  —  35^3       13  —  7^3 


10  2 

PROBLEM  IV. 

To  add  Surd  (Quantities  together. 

1.  Bring  all  fractions  to  a  common  denominator,  and  reduce 
the  quantities  to  their  simplest  terms,  as  in  the  last  problem. 
— 2.  Reduce  also  such  quantities  ds  have  unhke  indices  to 
other  equivalent  ones  having  a  common  index. — 3.  Then,  if 
the  surd  part  be  the  same  in  them  all,  annex  it  to  the  sum  of 
the  rational  parts,  with  the  sign  of  multiplication,  and  it  will 
give  the  total  sum  required. 

But  '}i  the  surd  part  be  not  the  same  in  all  the  quantities, 
they  can  only  Jm?  added  by  the  signs  -j-  and  —  . 
:  EXAMPLES. 

1 .  Required  to  add  ^  1 8  and  ^  32  together. 

First,  y  18  =  v^yX2  =  3^2;  and  ^^32  =  ^^16X2=  4^2  : 
Then,  3^2  +  4^2  =  (3  -f  4)^2  =  7^2  —  sum  required. 

2.  It  is  required  to  add  \/  375,  and  y  192  together. 

First,  ^3^75=1/  125X3=53/3;  and^/  192=3/64X3=43/3; 
Then,  5  ^3  -f  4^'  3  —  (5  4-  4)^3  -}-9  ^'3  =  sum  required. 

3.  Required! 


SURDS.  20p 

3.  Required  the  sum  of  y  27  and  ^  48.  Ans.  7  ^^  3. 

4    Required  the  sum  of  ^  50  and  ^  72.        Ans.  11^2. 

5.  Required  the  sum  of  ^  ^  and  y/  ^s- 

Ans.  4  ^  yV  or  A  \/  1^- 

6.  Required  the  sum  of  %/  56  and  %/  189.       Ans.  5  3/7. 

7.  Required  the  sum  of  y  ^  and  ^  -i-.  Ans.  \  %/  2. 

8.  Required  the  sum  of  3^a^b  and  5^  \6a^b.    » 

PROBLEM  V. 

To  find  the  Difference  of  Surd  Quantities. 

Prepare  the  quantities  the  same  way  as  in  the  last  rule  ; 
then  subtract  the  rational  parts,  and  to  the  remainder  annex 
the  common  surd,  for  the  difference  of  the  surds  required. 

But  if  the  quantities  have  no  common  surd,  they  can  only 
be  subtracted  by  means  of  the  sign — . 

EXAMPLES. 

1.  To  find  the  difference  between  ^  320  and  y  80. 


First,  ^  320  =  ^64  X  5=8^5  ;  and^80=y  16X5=4^5, 
Then  8^5  —  4-^5  =  4^5  the  difference  sought 

2.  To  find  the  difference  between  %/  128  and  y  54. 
First,  ^  128  =  ^64X2=43/  2  ;  and^/  4  =  3/27X2~=  »/  2. 
Then  4  ^  2  —  3  3/  2  =  y  2,  the  difference  required. 

3.  Required  the  difference  of  ^75  and  ^  48.    Ans.  ^  3. 

4.  Required  the  difference  of  1/  256  and  1/  32.  Ans.  23/4. 

5.  Required  the  difference  of  ^  ^  and  ^  |.     Ans.  i  1/ 6. 

6.  Required  the  difference  of  3/  |  and  y  %'.  Ans.  ^j^Td, 

7.  Find  the  difference  of  ^24a3  62  and  ^54ab*. 

Ans.  (a- 26)  ^  (362_2a6)  ^  6a,. 

PROBLEM  VL 

To  Multiply  Surd  Quantities  together. 

Reduce  the  surds  to  the  same  index,  if  necessary  ;  next 
multiply  the  rational  quantities  together,  and  the  surds  too-e- 
ther  ;  then  annex  the  one  product  to  the  other  for  the  whole 
product  required  ;  which  may  be  reduced  to  more  simple 
terms  if  necessary. 

,^  '  "  EXAMPLES. 


210  AI^GEBRA. 

EXAMPLES. 

1.  Required  to  find  the  product  of  4  ^  12,  and  3^2, 
Here,  4  X  3  X  v'  12X^/2=12 ^12 x'2=12^24=12y'4X6 

=  12  X  2  X  <v/  6  =  24  ^  6,  the  product  required. 

2.  Require  to  multiply  i  3/  i  by  |  »/  f . 
HereiXiVIXVI=TVXy3%-TVXl/if=TVX-}X^18 

=  jV  V  ^^j  *^^  product  required. 

3.  Required  the  product  of  3  y  2  and  2  ^  8.         Ans,  24. 

4.  Required  the  product  of  J-^4  and  1^12.        Ans.  i^6. 

5.  To  find  the  product  of  f^f  and  yVv^l-         Ans.  /o\/l^- 

6.  Required  the  product  of  23/14  and  33/4.     Ans.  123/7. 

,2.  i. 

7.  Required  the  product  of  2a^  an4  c^.  Ans.  2o2. 

i  3 

8.  Required  the  product  of  (a  -f  6)3  and  (a  -|-  6)*. 

9.  Required  the  product  of  2x  -{-  y/  b  and  2x  —  ^  b. 

10.  Required  the  product  of  (a  4-2v^6)2  and  (a — 2^6)-. 

11.  Required  the  product  of  2x  »  and  3x"*. 

12.  Required  the  product  of  4j;«  and  2^'» . 

PROBLEM  tn. 

To  Divide  one  Surd  Quantity  by  another. 

Reduce  the  surds  to  the  same  index,  if  necessary  ;  then 
take  the  quotient  of  the  rational  quantities,  and  annex  it  to 
the  quotient  of  the  surds,  and  it  will  give  the  wholf^  quotient 
required  ;  which  may  be  reduced  to  more  simple  terms  if 
requisite, 

EXAMPLES. 

1.  Required  to  dinde  6  v^  96  by  3  y  8. 

Here  6  —  3  .  ^  (96  ~  8)  =2^  12=2^^(4X3) =2X2^/3. 
=  4^3,  the  quotient  required. 

2.  Required  to  divide  123/280  by  33/6. 

Pere  12  ^  3  =  4,  and  280  -j-  5  =  66  =  8X7  =  23  .  7  ; 
Therefore  4  X  2  X  ^7  =  8^7,  is  the  quotient  required. 

3.  Let 


SURDS.  «11 

3.  Let  4  ;;/  50  be  divided  by  2  v'  6.  Ans.  2  ^  10, 

4.  Let  6  3/  100  be  divided  by  3  V'  ^-  Aas.  2  y  20. 

5.  Let  f  y  J^-  be  divided  by  f   ^  f .  Ans.  yV  v^  ^• 

6.  Let  f  y  ^j  be  divided  by  f  ^  f .  Ans,  ^  ^  30. 
•7.  Let  A  ^a,  or  f^t-,  be  divided  by  §a^.  Ans.  f a^. 

i.  3l 

8.  Let  a"  be  divided  by  a^. 

9.  To  divide  3a«  by  4a"». 


PROBLEM  VUI. 


To  Involve  or  Raise  Surd  Quantities  to  any  Power, 

Raise  botb  the  rational  part  and  the  surd  part.     Or  mul 
tiply  the  index  of  the  quantity  by  the  index  of  the  power  to 
which  it  is  to  be  raised,  and  to  the  result  annex  the  power  of 
the  rational  parts,  which  will  give  the  power'required. 


EXAMPLES. 

1.  Required  to  find  tbe  square  of  |a2. 

thirst,  (f )2  =  I  XJ  =  tV'  and  {a^y  ==  a^  ^  ^  =  a^  =.a. 
Therefore  (fa^)^  =  i\a,  is  the  square  required. 

3. 

2.  Required  to  find  the  square  of  ^a^ 
First,  i  X  i  =  i  and  (a^  )3  =  a»  =  a  3/  « ; 
Therefore  (ict^)^  =  la^/a  is  the  square  required. 

3.  Required  to  find  the  cube  of  |  ^  6  or  |  X  6^' 
First,  (1)3  =  I  X  f  X  I  =  ^\  and  (6^)3  =6^  =  6^6. 
Theref  (1^6)3  = /^  x6y/6=  V  a/ 6,  the  cube  required. 

4.  Required  the  square  of  2  ^  2.  Ans   4^4. 

6.  Required  the  cube  of  3^,  or  ^  3.  Ans.  3  ^  3. 
G,  Required. the  3d  power  of  i  ^  3.                 Ans.  }  ^  3. 

7.  Required  to  find  th©  4th  power  of  ^  ^2.  Ans.  }. 

8.  Required 


212  ALGEBRA. 

i. 

8.  Required  to  find  the  mth  power  of  a". 

9.  Required  to  find  the  square  of  2  -f  ^  3. 

PROBLEM  IX. 

To  Evolve  or  Extract  the  Roots  of  Surd  Quantities*. 

Extract  both  the  rational  part  and  the  surd  part.  Or  di- 
vide the  index  of  the  given  quantity  by  the  index  of  the  root 
to  be  extracted  ;  then  to  the  result  annex  the  root  of  the  ra- 
laonal  part,  which  will  give  the  root  required. 

EXAMPLES. 

1.  Kequired  to  find  the  square  root  of  16-^6. 
irirst,  ^  16  =  4,  and  {f)^  =ze^^^  =  6^  ; 

theref.  (16  y  6)2  =  4  .  6*v=  4  y  6,  is  the  sq.  root  required^. 

2.  Required  to  find  the  cube  root  of  -^-^  y/  3. 
First,  %/  ^V  =  h  and  W.^f  =  32  -^  ^  =  36  ; 

theref.  (2V  ^  3)^  =  i  .  3^  =  i  ^  3,  is  the  cube  root  required. 

5.  Required  the  square  root  of  G^.  Ans.  6  y^  6. 
4.  Required  the  cube  root  of  \a'^h.                  Ans.  \a  ^6. 

6.  Required  the  4th  root  of  16a2.  Ans.  2  ^  a, 

6.  Required  to  find  the  mth  root  of  x^. 

7.  Required  the  square  root  of  a^  ^  ^a  ^h  ■\-  96. 


*  The  square  root  of  a  binomial  or  residual  surd,  a  +  *»  <>i'  ^  -~  ^ 
Hiay  be  found  thus  :  Take  ^  a^  -b'^  ^  c; 
— — .  rt  +  C  «  — c 

then  v/  a  -}-  6  x=  x/ '  +  \/ > 

2                   2 
— —            rt-fc-c            a—c 
and  v^  a  — A  =  v/——  —  \/ • 

Thus  the  square  root  of4  +  2v^3  =  l-f-v^3; 

and  the  square  root  of  6  •—  2  ^^  5  —  ^  5  —  1. 

But  for  the  cube,  or  any  higher  root,  no  general  rule  is  known. 

INFINITE 


SURDS.  213 

INFINITE  SERIES. 

An  Infinite  Series  is  formed  either  from  division,  dividing 
hy  a  comj)ound  divisor,  or  by  extracting  the  root  of  a  com- 
pound surd  quantity  ;  and  is  such  as,  being  continued,  would 
run  on  infinitely,  in  the  manner  of  a  continued  decimal  frac- 
tion. 

But,  by  obtaining  a  few  of  the  first  terms,  the  law  of  thfr^ 
pros^ression  will  be  manifest ;  so  that  the  series  may  thence  be 
continued,  without  actually  performing  the  whole  operation. 

PROBLEM  I. 
To  Reduce  Fractional  Quantities  into  Infinite  Series  by  Division. 

Divide  the  numerator  by  the  denominator,  as  in  common 
division  ;  then  the  operation,  continued  as  far  as  may  be 
thought  necessary,  will  give  the  infinite  series  required. 

EXAMPLES. 

2ab 

1.  To  change into  an  infinite  series. 

a  4-  6 

«  +  6)  2a6  .  .  (  26 1 +  &c. 

a         a2       a3 
2o6  +  262 

.-  262 

26» 
«  263 


268 


263       26* 
a         a3 


a3 
26*        26s 

a^         o3 


,  &c. 

a' 

'"  2.  Let 


214  ALGEBRA. 

2.  Let be  changed  into  an  infinite  series. 

1  —  a 

1  —  a)  1  (  1  +  a  +  a2  +  a3  4-  o*  +  &c. 

1  --  a 


a2 


a2 

a2  -C3 


a3«a4 


h 

3. 

Expand            ■  into  an  infinite  series. 

h               c 

C2           C3 

Ans.  — X<1 1 

+  &C.) 

a                a 

a2      a» 

a 

4. 

Expand into  an  infinite  series. 

a  —  b 

h 

is             fc3 

Ans.  1-1 1- 

-  +  -  +  &C. 

a  . 

a2       a3 

1  -  a; 

6. 

Expand into  an  infinite  series. 

1  -f  re 

Ans.  1  —  2a;  -f  2a;2— .gx^  4-  2x*y  &c. 

6.  Expand  ■"  into  an  infinite  series. 

26       362     463 

'  Ans.  1 1 ,&c 

a         aa       o=» 
1 

7.  Expand =^,  into  an  infinite  series. 

l-fl 

PJROBLEM  II. 

To  Reduce  a  Compound  Sufd  into  an  Infinite  Series. 
Extract  the  root  as  in  common  arithmetic  ;  then  the  ope- 
ration, continued  as  far  as  may  be  thought  necessary,  will  give 
the  series  required.  But  this  method  is  chiefly  of  use  in  ex- 
tracting the  square  root,  the  operation  being  too  tedious  for 
the  higher  powers.  EXAMPLES. 


INFINITE  SERIES.  f  If 

EXAMPLES. 

K  Extract  the  root  of  a^  —  x'  in  an  infinite  series. 
x^        X*         X*  6x3 

a»  — ar2  (^ —  &c. 

ia       8o3      I6a«       1280' 
a2 


x^ 

2a 

2a 

^x» 

4a3 

X* 

X8 

X* 

2a  < 

a 

8o3 

)- 

4a3 

X9 

x» 

' + 

— + 

JC-3 

X4 

4a2 

Sa* 

64a» 

ore 

a;' 

B 

2a- 

-.    . 

.   -    — 

&c.) 

.; 

_  —— 

— 

a 

4o3 

8a* 

64a< 

x^ 

S 

-f 

&c. 

8a* 

16a« 

6a:« 

%t 

' 

64a6 

&c. 

2.  Expand  ^Z  1  +  1  =  \/  ^>  into  an  infinite  series. 

Ans  1  +  i-i  +  i^^jh  &c. 

3.  Expand  y^  1  —  1  into  an  infinite  series., 

Ans.  1-±-.i-J--t|3.  &c. 

4.  Expand  ^  a'^  +  x  into  an  infinite  series. 

5.  Expand  ^  a^  —  26a;  -  x^  to  an  infinite  series. 

PROBLEM  III. 

To  Extract  any  Root  of  a  BinomifU :  or  to  Reduce  a  Binomial 

Surd  into  an  Infinite  Series. 

This  will  be.doue  by  substituting  the  particular  letters  of 
the  binomial,  with  their  proper  signs,  in  the  following  general 
theorem  or  formula,  viz. 

m         mm  m  —  n  m—2n 

(P  +  Pft)    *»  =  P  n  -I A^  -I Bft  H C€l  4-  &C. 

n  2n  3»  and 


216  ALGEBRA. 

and  it  will  give  the  root  required  :  observing  that  p  denotei 
the  first  term,  q  the  second  term  divided  by  the  first,  ^  the 
index  of  the  power  or  root ;  and  a,  b,  c,  d,  &,c.,  denote  the 
several  foregoing  terms  with  their  proper  signs. 

EXAMPLES. 

1.  To  extract  the  sq.  root  of  a^  +  ^",  in  an  infinite  series. 

6«  m  1 

Here  p  =  a^ ,  ci  =  —  and  —  =  —  :  therefore 

m  m  X 

jp  -  =  (o^)  -  =  (aa)  2  =  a  =  A,  the  1st  term  of  the  series. 

m  b''       b^ 

—  AQ=|^  X  aX  —  =  — =B,  the  2d  term: 

n  a^       2a 

m—n  1—2     62     fc2  54 

B^  X X— X—  =  — =  c,  the  3d  term, 

2n  4       2a     eta  2.4a3 

m-^2n  1—4  b^         b^  3b^ 

c^  = X X—  = =  D  the  4tli. 

3fi  6  2.4a3     a3        2.4.6a5 

.62         h  3.6  e 

Hence  a  -f- j —  &c.  or 

2a      2.4a3       2.4  6a^ 
62         5*  66  568 

a  -{ 1 &c.  is  the  series  required. 

2a       8a3        16a«      1280^ 

1  -^ 

2.  To  find  the  value  of ,  or  its  equal  (a  -  a;)-2 ,  in  an 

infinite  series*.  (a~x) 


♦  N'ote,  To  facilitate  the  af  plication  of  the  rule  to  fractional  ex- 
amples, it  is  proper  to  observe,  that  any  surd  may  be  taken  from  the 
denominator  of  a  fraction  and  placed  in  the  numerator,  and  vice  versa^ 
by  only  changing  the  sign  of  its  index.    Thus, 
1  1 

— =IX:j^'  or  only  jr  2  ;  apd .  ■»  1  X  (o  +  6)  ~*  or 

^2  (a  +  ^)2  ^ 

aa  ^*        ,        I 

(a  +6)~2  ;  and— ^a^{<i-\-xy~^;  and — —a:^x^^  ;  als» 

,        {a  -f  :r)2  x^ 

(fl2  4.  a?2)2  ^  -X 

,  =  (a2  +  J»)2  X  (a2  -,ar2)  ^  ,  g^^, 

ga2  _  x2  )2 

Here 


INFINITE  SERIES.  2f7 

—a:  m       —2 

Herep  =  a,  a  = =  — a"'x,  and— = =—2  ;  theref. 

a  n         1 

P  ^  =  ro^""2  r=  =  A,  th  e  1st  term  of  the  series. 

«       ^  '  o2 

—  A€i  =  —  2X--X =  —  =  2a    3  ;i^  =  ^^  the  2d  term. 

n                          a*          a         a=* 
m  — n                      2x       "X       3x2 
Ba  =  — I  X  — X = =  3a""*a:2  =  c,  the  3d. 

2ra  a3         a  a* 

m— 2n  3j;2       —o:      ^x"^ 

cft  =  — f  X X = =  4a~5  j:3  =  ©. 

3n  a*         a  a* 

Hence  ar^-\'^a"^x-\-'^a~'^x^  -f-  ^a~^x^  -{-  &c,  or 

1        2a:       3x3        4j;3        5^4 

1 1 1 1 &c,  is  the  series  required. 

a2        a^        a*         a^         a^ 
a2 

3.  To  find  the  value  of ,  in  an  infinite  series. 

a*^x 

x^       x^       x^ 

Ans.  o  +  x-l 1 1 &c. 

a        a2        a» 

4.  To  expand  v^,      .       .  or  ,      .       .  J"  in  a  series. 

1         ar2        3x4        5^6 

Ans.  —  — 1 &c. 

a       2a3       8a5         16a'' 
a2 

5.  To  expand  — ' in  an  infinite  series. 

(a-6)2 

26       362       453       5J4 

Ans.  1  -I 1 h 1 &c, 

a         a^         a2         a* 

^.  To  expand -^a2  — x2  or  (a2  —  x^y  in  a  series. 

x^      X*       x«       6z« 

Ans.  a &c. 

2a      8a  3      ISa^      128a7 

7.  Find  the  value  of  y  (a^-  63)  or  (a3  —  63)3  in  a  series. 

63       6«        669 

Ans.  a ' — —1 < &.C. 

3a2     9a«       81a» 

5.  To  find  the  value  of  ly  (a^  -|-x*  )  or  (a*  4-^*)^  in  a  series. 
x^       2x»»         6x»« 

Ans.  a  H [-  . &c. 

5a*      SSa"        125a»* 
Vol.  I.  29  .  9.  To 


218  ALGEBRA. 

9.  To  find  the  square  root  of in  an  infinite  series. 

b  X^  073 

Ans.  1 1 &c. 

a         2a2     2a3 

^     10.  Find  the  cube  root  of ,  in  a  series. 

3a3        9a^         81«» 


ARITHMETICAL  PROPORTION^. 


Arithmetical  Proportion  is  the  relation  between  tw# 
»auHibers  with  respect  to  their  difference. 

Four  quantities  are  in  Arithmetical  Proportion,  when  the 
difference  between  the  first  and  second  is  equal  to  the  dif- 
ference, between  the  third  and  fourth.  Thus,  4,  6,  7,  9,  and 
a,  a  -{-  d^btb  X  d  are  in  arithmetical  proportion. 

Arithmetical  Progression  is  when  a  series  of  quantities 
have  all  the  same  common  difference,  or  when  they  either 
increase  or  decrease  by  the  same  common  difference.  Thus 
2,  4,  6,  8,  10,  12,  &c.  are  in  arithmetical  progression  having 
the  common  difference  2  ;  and  a,  a  -{-  c?,  a  -f-  2<f,  a  -f  3d, 
a  -\-  4d,  a  -{-  Bd,  &c.  are  series  in  arithmetical  progression, 
the  common  difference  being  d. 

The  most  useful  part  of  arithmetical  proportion  is  contained 
in  the  following  theorems  : 

1.  When  four  quantities  are  in  Arithmetical  Proportion, 
the  sum  of  the  two  extremes  is  equal  to  the  sum  of  the  two 
means.  Thus,  in  the  arithmetical  4,  6,  7,  9,  the  sum  4  + 
9  =  64-7=13:  and  in  the  arithmeticals  a,  a  -{-  d,  fc,  6  -{-  </, 
the  sum  a  +  6  -|-  d  =  a  -f-  6  -{-  d« 

2.  In  any  continued  arithmetical  progression,  the  sum  of 
the  two  extremes  is  equal  to  the  sum  of  any  two  terms  at  an 
equal  distance  from  them. 

Thus, 


ARITHMETICAL  PROPORTION.  219 

Thus,  if  the  series  be  1,  3,  6,  7,  9,  11,  &c. 
Then  1  +  11  =  3 -f  9  =  5 -f  7  =  12. 

3.  The  last  term  of  any  increasing  arithmetical  series,  is 
equal  to  the  first  term  increased  by  the  product  of  the  com- 
mon difference  multiplied  by  the  number  of  terms  less  one  ; 
but  m  a  decreasing  series,  the  last  term  is  equal  to  the  first 
term  lessened  by  the  said  product. 

Thus,  the  20th  term  of  the  series,  1,  3,  6,  7,  9,  &c.  is  ~ 
1  4-  2  (20-1)  =  1  +2  X  19  =  1  +38  =  39. 

And  the  nth  term  of  c,  a  — c?,  a-- 2d,  a— 3rf,  a  -  4d,  &c.  is 
=  a-(n— 1)  X«?  =  a— (n  — l)d. 

4.  The  sum  of  all  the  terras  in  any  series  in  arithmetical 
progression,  is  equal  to  half  the  sum  of  the  two  extremes 
multiplied  by  the  number  of  terms. 

Thus,  the  sum  of  1,3,  6,  7,  9,  &c.  continued  to  the  10th 
(1  -I-  19)  X  10       20  X  10 
term,  is  =  ^^ ^' — ^ —  =  10  X  10  =  100. 

And  the  sum  of  n  terms  of  a,  a  -f-  d,  a  4*  2c?,  a  -f-  3d,  to 
n 
a  -{-  md,  is  =  (a  -|-  a  -{-  tnd)  .  —  =  (^  +  i  md)n. 

'    2 


EXAMPLES  FOR  PRACTICE. 


1.  The  first  term  of  an  increasing  arithmetical  series  is  I, 
the  common  difference  2,  and  the  number  of  terms  21  ;  re- 
quired the  sum  of  the  series  ? 

First,  1  -}-  2  X  20  =  1  -f  40  =  41,  is  the  last  term. 
1  +  41 
'  Then  — z —   X  20  =  21  X  20  =  420,  the  sum  required. 

2.  The  first  term  of  a  decreasing  arithmetical  series  is  199, 
the  common  difference  3,  and  the  number  of  terms  67  ;  re- 
quired the  sum  of  the  series  ? 

First,  199— 3  .  66  =  199—198  =  1,  is  the  last  term. 
Then  i?^  X  67  =  100   X  67  =  6700,  the  sum  re- 
quired. 

3.  To  find  the  sum  of  100  terms  of  the  natural  numbers 
1,  2,  3,  4,  6,  6,  &c.  Ans   6050. 

4.  Required 


22^  ALGEBRA. 

4.  *  Required  the  sum  of  99  terms  of  the  odd  numbers 
1,3,  6,7,  9,  &c.  Ans.  9811. 

6.  The  first  term  of  a  decreasing  arithmetical  series  is  10, 
the  common  difiference  ^,  and  the  number  of  terms  21  ;  re- 
quired the  sum  of  the  series  ?  Ans.  140. 

6.  One  hundred  stones  being  placed  on  the  ground,  in  a 
straight  line,  at  the  distance  of  2  yards  Irom  each  other  ;  how 
far  will  a  person  travel,  who  shall  bring  them  one  by  one  to 
a  basket,  which  is  placed  2  yards  from  the  first  stone  ? 

Ans  1 1  miles  and  840  yards. 


APPLICATION  OF  ARITHMETICAL  PROGRESSION 
TO  MILITARY  AFFAIRS. 

QUESTION  I. 

A  Triangular  Battalion,!  consisting  of    thirty  ranks,  in 
which  the  first  rank  is  formed  of  one  man  only,  the  second 

of  3 


*  The  sum  of  any  number  (n)  of  terms  of  the  arithmetical  series  of 
odd  number  1,  3,  5,  7,  9,  &c.  is  equal  to  the  square  («3  )  of  that  num- 
ber.    1  hat  is, 
If  1,  3,    5,    7,     9,    &c.  be  the  numbers,  then  will 

12,2^,  33,  42,  53,  &c.  be  the  sums  of  1,2,  3,  &c.  terms. 
Thus,  0  -|-  1  =     1  or  12 ,  the  sum  of  1  term, 
14-3=    4  or  22 ,  the  sum  of  2  terms, 
4  +  5  =     9  or  32 ,  the  sum  of  3  terms, 
9  Hh  7  =  16  or  42 ,  the  sum  of  4  terms,  &€• 
For,  by  the  3d  theorem,  1+2  (n— 1)  =  1  +  2n— 2  =  2n— 1  is  the 
last  term,  when  the  number  of  terms  is  n ,-  to  this  last  term  2n  —  1, 
add  the  first  term  1,  gives  2n  the  sum  of  the  extremes,  or  n  half  the 
sum  of  the  extremes  ;  then,  by  the  4ih  theorem,  ?iXn  t=  w2  is  the  sum 
of  all  the  terms.     Hence  it  ai)pears  in  general,  that  half  the  sum  of 
the  extremes,  is  always  the  same  as  the  number  of  the  terms  n  ;  and 
that  the  sum  of  all  the  terms,  is  the  same  as  the  square  of  the  same 
number,  n2. 

See  more  on  Arithmetical  Proportion  in  the  Arithmetic,  p.  Ill' 

f  By  triangular  battalion,  is  to  be  understood,   a  body  of  troops, 
ranged  in  the  form  of  a  triangle,  in  which  the  ranks  exceed  each 

other 


ARITHMETICAL  PROGRESSION.  221 

of  3  the'third  of  5  and  so  on  :  What  is  the  strength  of  such  a 
triaigular  battalion  ?  Answer,  900  men. 


QUESTION  n. 

A  detachment  having  12  successive  days  to  march,  with  or- 
ders to  advance  the  first  day  only  2  leagues,  the  second  S^,  and 
so  on  increasing  1^  league  each  day's  march  :  What  is  the 
length  of  the  whole  march,  and  what  is  the  last  day's  march  ? 

Answer,  the  last  day's  march  is  18^  leagues,  and  123  leagu€fs 
is  the  length  of  the  whole  march. 


QUESTION  HL 


A  brigade  of  sappers,*  having  carried  on  15  yards  of  sap 
the  first  night,  the  second  only  13  yards,  and  so  on,  decreasing 
2  yards  every  night,  till  at  last  they  carried  on  in  one  night 
only  3  yards  :  What  is  the  number  of  nights  they  were  em- 
ployed ;  and  what  is  the  whole  length  of  the  sap  ? 

Answer,  they  were  employed  7  nights,  and  the  length  of  the 
whole  sap  was  63  yards. 


other  by  an  equal  number  of  men  ;  if  the  first  rank  consist  of  one  man, 
only,  and  the  diiference  between  the  ranks  be  also  l,then  its  form  is 
that  of  an  equilateral  triangle  ;  and  when  the  difference  between  the 
ranks  is  more  than  1,  its  form  may  then  be  an  isosceles  or  scalene  tri- 
angle. The  practice  of  forming  troops  in  this  order,  which  is  now 
laid  aside,  was  formerly  held  in  greater  esteem  than  forming  them  in 
a  solid  square  as  admitting  of  a  greater  front,  especially  when  the 
troops  were  to  make  simply  a  stand  on  all  sides. 

*  A  brigade  of  sappers,  consists  generally  of  8  men  divided  equally 
into  two  parties.  While  one  of  these  parties  is  advancing  the  sap,  the 
other  is  furnishing  the  gabions,  fascines,  and  other  necessary  imple- 
ments, and  when  the  first  party  is  tired,  the  second  takes  its  place 
and  soon,  till  each  man  in  turn  has  been  at  the  head  of  the  sap.  A  sap 
is  a  small  ditch,  between  3  and  4  feet  in  breadth  and  depth  ;  and  is 
distinguished  from  the  trench  by  its  breadth  only,  the  trench  having 
between  10  and  15  feet  breadth.  As  an  encouragement  to  sappers, 
the  pay  for  all  the  work  carried  on  by  the  whole  brigade,  is  given  to 
the  survivors. 

QUESTIOX 


222  ALGEBRA. 

QUESTION  rV. 

A  number  of  gabions  *  being  given  to  be  placed  in  six 
ranks,  one  above  the  other,  in  such  a  manner  as  that  each 
rank  exceeding  one  another  equally,  the  first  may  consist  of 
4  gabions,  and  the  last  of  9  :  What  is  the  number  of  gabions 
in  the  six  ranks  ;  and  what  is  the  difference  between  each 
rank  ? 

Answer,  the  difference  between  the  ranks  will  be  1 ,  and 
the  number  of  gabions  in  the  six  ranks  will  be  39, 

QUESTION  V. 

Two  detachments,  distant  from  each  other  37  leagues,  and 
both  designing  to  occupy  an  advantageous  pest  equi-distant 
from  each  other's  camp,  set  out  at  different  times  ;  the  first 
detachment  increasing  every  day's  march  I  league  and  a  half, 
and  the  second  detachment  increasing  each  day's  march  2 
leagues  :  both  the  detachments  arrive  at  the  same  time  ;  the 
first  after  5  days'  march,  and  the  second  after  4  days'  march  : 
V/hat  is  the  number  of  leagues  marched  by  each  detachment 
each  day  ? 

The  progression  ^V,  2^3^,  Sf^,  5y\,  6j\y  answers  the  con* 
ditions  of  the  first  detachment  and  the  progression  If,  3f,  £|, 
7f ,  answers  the  conditions  of  the  second  detachment. 

QUESTION  VI. 

A  deserter,  in  his  flight,  travelling  at  the  rate  of  8  leagues  a 
day  ;  and  a  detachment  of  dragoons  being  sent  after  him  with 
orders  to  march  the  first  day  only  2  leagues,  the  second  5 
leagues,  the  third  8  leagues,  and  so  on  :  What  is  the  number  of 
days  necessary  for  the  detachment  to  overtake  the  deserter, 
and  what  will  be  the  number  of  leagues  marched  before  he  is 
overtaken? 

Answer,  5  days  are  necessary  to  overtake  him  ;  and  conse- 
quently 40  leilgues  will  be  the  extent  of  the  march. 


*  Gabions  are  baskets,  open  at  both  ends,  made  of  ozier  twigs,  and 
of  a  cylindrical  form  ;  those  made  use  of  at  the  trenches  are  2  feet 
wide,  and  about  3  feet  high  ;  which,  being  filled  with  earth,  serve  as 
a  shelter  from  the  enemy's  fire  ;  and  those  made  use  of  to  construct 
batteries,  are  generally  higher  and  broader.  There  is  another  sort  of 
gabion  made  use  of  to  raise  a  low  parapet ;  its  height  is  from  1  to  2 
feet,  and  I  foot  wide  at  top,  but  somewhat  less  at  bottom,  to  give  room 
for  placing  the  muzzel  of  a  firelock  between  them  j  these  gabions 
serve  instead  of  sand  bags.  A  sand  bag  is  generally  made  to  contain 
about  a  cubical  foot  of  earth. 

QUESTION 


PILING  OP  BALLS, 


22? 


QXJESTION  Vn, 

A  convoy*  distant  35  leagues,  having  orders  to  join  its 
camp,  and  to  march  at  the  rate  of  5  leagues  per  day  ;  its 
escort  departing  at  the  same  time,  with  orders  to  march  the 
first  day  only  half  a  league,  and  the  last  day  9^  leagues  ;  and 
both  the  escort  and  convoy  arriving  at  the  same  time  :  At 
what  distance  is  the  escort  from  the  convoy  at  the  end  of  each 
march  ? 

OF  COMPUTING  SHOT  OR  SHELLS  IN  A  FINISHED  PILE. 


Shot  and  Shells   are   generally  piled   in  three  different 
forms,  called  triangular,  square,  or  oblong  piles,  according 
as  their  base  is  either  a  triangle,  a  square,  or  a  rectangle. 
Fig,  1.       C  G^  Fig,  2, 


H 


B  E 

ABCD,  fig.  1,  is  a  triangular  pile, 
EFGH,  fig.  2,  is  a  square  pile.  . 
..       E 


F  D    ' 

ABCDEF,  fig.  3,  is  an  oblong  pile. 


*  By  convoy  is  generally  meant  a  supply  of  ammunition  or  provi- 
sions, conveyed  to  a  town  or  army.  The  body  of  men  that  guard 
this  supply  is  called  escort. 


A  triangular 


224'  ALGEBRA. 

A  triangular  pile  is  formed  by  the  continual  laying  of  trian- 
gular horizontal  courses  of  shot  one  above  another,  in  such  a 
manner,  as  that  the  sides  of  these  courses,  called  rows, 
decrease  by  unity  from  the  bottom  row  to  the  top  row,  which 
ends  always  in  1  shot. 

A  square  pile  is  formed  by  the  continual  laying  of  square 
horizontal  courses  of  shot  one  above  another,  in  such  a  man- 
ner, as  that  the  sides  of  these  courses  decrease  by  unity  from 
the  bottom  to  the  top  row,  which  ends  also  in  1  shot. 

In  the  triangular  and  the  square  piles,  the  sides  or  faces 
being  equilateral  triangles,  the  shot  contained  in  those  faces 
form  an  arithmetical  progression,  having  for  first  term  unity^ 
and  for  last  term  and  number  of  terms,  the  shot  contained  in 
the  bottom  row  ;  for  the  number  of  horizontal  rows,  or  the 
number  counted  on  one  of  the  angles  from  the  bottom  to  the 
top,  is  always  equal  to  those  counted  on  one  side  in  the  bot- 
tom :  the  sides  or  faces  in  either  the  triangular  or  square  piles, 
are  called  arithmetical  triangles  ;  and  the  numbers  contained 
in  these,  are  called  triangular  numbers  :  abc,  fig.  1,  efg,  fig. 
2,  are  arithmetical  triangles. 

The  oblong  pile  may  be  conceived  as  formed  from  the 
isquare  pile  abcd  :  to  one  side  or  face  of  which,  as  ad,  a 
number  of  arithmetical  triangles  equal  to  the  face  have  beea 
added  :  and  the  number  of  arithmetical  triangles  added  to  the 
square  pile,  by  means  of  which  the  oblong  pile  is  formed,  is 
always  one  less  than  the  shot  in  the  top  row  ;  or,  which  is  the 
same,  equal  to  the  difierence  between  the  bottom  row  of  the 
greater  side  and  that  of  the  lesser. 

QUESTION  VIIL 

To  find  the  shot  in  the  triangular  pile  abcd,  fig.  1,  the 
bottom  row  ab  consisting  of  8  shot. 


SOLUTION. 

'The  proposed  pile  consisting  of  8  horizontal  courses,  each 
of  which  forms  an  equilateral  triangle  ;  that  is,  the  shot  con- 
tained in  these  being  in  an  arithmetical  progression,  of  which 
the  first  and  last  term,  as  also  the  number  of  terms,  are 
known  ;  it  follows,  that  the  sum  of  these  particular  courses, 
or  of  the  8  progressions,  will  be  the  shot  contained  in  the 
proposed  pile  ;  then 

The 


/;  ^  >  ^ 


PILING  OF  BALLS. 


225 


The  shot  of  the  first  or  lower  } 
triangular  course  will  be       ^ 

the  second     _   - 

the  third  -  .  -  . 

the  fourth  -  -  -  • 

the  fifth  .  -  .  . 

the  sixth  -  -  -    -     - 

the  seventh  .  - 

the  eighth  -  -  -  - 


8  +   1    >4^4b  =  36 
7   -f    1    X    3J^  =  2« 


6  -f   1    X   3     =21 


6-1-1    X    2i  =  !• 
4  4-1X2     =1© 


3  4-   1    X    U 


2  -f   1    X    1     =3 


1   -1-    1    X 


=     1 


Total     -       120  shot 
in  the  pile  proposed. 
QUESTION  IX. 

To  find  the  shot  of  the  square  pile  efgh,  fig.  2,  the  bottom 
row  EF  consisting  of  8  shot. 

SOLUTION. 

The  bottom  row  containing  8  shot,  the  second  only  7  ; 
that  is,  the  rows  forming  the  progression,  8,  7,  6,  5,  4,  3,  2,  1, 
in  which  each  of  the  terms  being  the  square  root  of  Ihe  shot 
contained  in  each  separate  square  course  employed  in  forming 
the  square  pile  ;  it  follows,  that  the  sum  of  the  squares  of 
these  roots  will  be  the  shot  required  :  and  the  sum  of  the 
squares  of  8,  7,  6,  6,  4,  3,  2,  1,  being  204,  expresses  the  shot 
in  the  proposed  pile. 

QUESTION  X. 

To  find  the  shot  of  the  oblong  pile  abcdef,  fig.  3  ;  in 
which  BF  =  16,  and  bc  =  7. 

SOLUTION. 

The  oblong  pile  proposed,   consisting  of  the  square  pile 
ABCD,   whose  bottom  row   is   7  shot  ;  besides  9  arithmetical 
triangles  or  progressions,  in  which  the  first  and  last  term,  as 
also  the  number  of  terms,  are  known  ;  it  follows,  that, 
if  to  the  contents  of  the  square  pile  -         140 

we  add  the  sum  of  the  9th  progression         -         252 

their  total  gives  the  contents  required  -        392  shot. 

REMARK  I. 

The  shot  in  the  triangular  and  the  square  piles,  as  also 
ihe  shot  in  each  horizontal  course,  may  at  once  be  ascer- 

VoL.  I.  30  tained 


226 


ALGEBRA. 


tained  by  the  following  table  :  the  vertical  column  a,  con- 
taifis  the  s^ot>  in  the  bottom  row,  horn  I  to  20  iDclu§ive  ; 
the  robmiu  b  contains  the  triangular  numbers,  or  nuuiher 
of  each  course  ;  the  column  c  contains  the  sum  of  the 
triangular  numbers,  that  is,  the  sh>>t  contained  in  a  triangular 
pile,  con;mor.ly  called  pyramidal  nunibers  ;  the  column  d 
contains  the  square  of  the  numbers  of  the  column  a,  that  is, 
the  shot  coritaijK  d  in  each  square  horizontal  course  ;  and  the 
column  E  contams  the  sum  of  these  squares  or  shot  in  a 
square  pile. 


c 

B 

A 

D 

E 

S..U.  .  i-- 

pyramidal 
numbers. 

Triangular 
numbers. 

Natural 
numbers. 

•Square  ol 
ihe  natural 
numbers. 

'  hese 

tquyre 

nur»  berg. 

1 

1 

1- 

1 

1 

4 

3 

2 

4 

5 

,        10 

6 

3 

9 

14 

20 

10 

4 

16 

30 

35 

15 

5 

25 

55 

66 

21 

6 

36 

91 

84 

28 

7 

49 

140 

120 

36 

8 

64 

204 

165 

45 

9 

81 

285 

220 

55 

10 

100 

385 

286 

66 

li 

121 

506 

364 

78 

12 

144 

650 

455 

91 

13 

169 

819 

560 

105 

14 

K;6 

1015 

680 

120 

15 

225 

1240 

816 

136 

16 

256 

1496 

969 

153 

17 

289 

1785 

1140 

171 

18 

324 

2109 

1330 

190 

19 

361 

2470 

1540 

210 

20 

400 

28*/© . 

Thus,  the  bottom  rrw  in  a  triangular  pile,  consisting  of 
9  shot,  the  contents  will  he  165  ;  and  when  of  9  in  the  square 
pile,  285. — !n  the  same  manner,  the  contents  either  of  a 
scuare  or  tri?-nj.nilar  pile  being  given,  the  shot  in  the  bottom 
row  may  be  easily  ascertained. 

The  contents  of  any  oblong  pile  by  the  preceding  table 
may  be  also  with  little  trouble  ascertained,  the  less  side  not 
exceedir  g  20  shot,  nor  the  difference  between  the  less  and 
the  greater  side  SO.     Thus,  to  find  the  shot  in  an  oblong  pile, 

the 


PILING  OF  BALLS.  S2T 

the  loss  side  being  15,  and  the  greater  35,  we  are  first  to 
find  tlie  contents  of  the  sqaare  pile  by  means  of  which  the 
oblong  pile  may  he  conceived  to  be  formed  ;  that  is,  we  are 
to  find  the  contents  of  a  square  pile,  whose  bottom  row  is 
15  shot  V  whicii  being  1240,  we  are,  secondly,  to  add  these 
1240  to  the  product  2400  of  the  triangular  number  120, 
answering  to  15,  the  number  expressing  the  bottom  row  of  the 
arithmetical  triangle,  multiplied  by  20,  the  number  of  those 
triangles  ;  and  their  sum,  being  3G40,  expresses  the  number 
of  shot  in  the  proposed  oblong  pile. 


REMARK  II. 

The  following  algebraical  expressions,  deduced  from  the 
investigations  of  the  sums  of  the  powers  of  numbers  in 
arithmetical  progression,  which  are  seen  upon  many  gunners' 
callipers*,  serve  to  compute  with  ease  and  expedition  the  shot 
or  shells  in  any  pile. 
That  serving  to  compute  any  triangular  }  n  ■{-  2  x  »  -r   ^  X  n 

pil«,  is  represented  by  ^  6 

That  serving  to  compute  any  square  (  n  -f~  1  X  2n  -}-  I  X  n 

pile,  is  represented  by  (  & 

In  each  of  these,  the  letter  n  represents  the  number  in  the 
bottom  row  ;  hence,  in  a  triangular  pile,  the  number  in  the 
bottom  row  being  30  ;  then  this  pile  will  he  30  +  2  X  30  -^  1 
X  y  =  4960  shot  or  shells.  In  a  square  pile,  the  number 
in    the   bottom   row    being   ali^o  30  ;  then   this   pile  will   be 

30  +  1  X  60  f  1  X  Y  =  ^-^-^-^  shot  or  shells 
.  That  serving  to  compute  any  oblong  pile,  is  represented  bj 

2n  +  1   4-''¥n  X  n'i^l    X  ?i 

— ,  in  which  the  letter  n  denotes 


*  Callipers  ar«  large  compasses,  with  bowed  shank?,  serving  to  take 
the  diameters  of  convex  and  concave  bodies.  Tiie  gunners*  callipers 
cons. St  of  two  thin  rules  op  plates,  which  are  moveible  quite  round 
a  joint,  by  the  plates  folding  one  over  ihe  other  :  the  length  of  each 
rule  or  piate  is  6  inches,  the  breadth  about  1  mch.  It  is  usual  to  re- 
prese.u,  on  the  platen,  a  variety  of  scales,  tables,  proportions,  &c. 
such  as  are  otesme  I  useful  to  be  known  by  persons  employed  about 
artillery  ;  but  except  the  meaaurm^  of  the  caliber  of  shot  and  cannon, 
and  the  measuring  of  saliant  and  re-entering  an^^les,  none  of  the  'it  ti- 
des, with  which  the  callipers  are  usually  rilled,  are  essential  to  that 
inetrument. 

the 


^28  ALGEBRA. 

the  number  of  courses,  and  the  letter  m  the  number  of  shot, 
less  one,  in  the  top  row  :  hence,  in  an  oblong  pile  the  num- 
ber of  courses  being;  30,  and  the  top  row  31  ;  this  pile  will 
be  60^4-  1  +"90  X  30 -f  1  +  V  =  ^3405  shot  or  shells. 


^GEOMETRICAL  PROPORTION. 

(geometrical  Proportion  contemplates  the  relation  of 
quantities  considered  as  to  what  part  or  what  multiple  one  is^ 
of  another,  or  how  often  one  contains,  or  is  contained  in, 
another.- — Of  two  quantities  compared  together,  the  ^rst  is 
called  the  Antecedent,  and  the  second  the  Consequent.  Their 
ratio  is  the  quotient  which  arises  from  dividing  the  one  by  the 
other. 

Four  Quantities  are  proportional,  when  the  two  couplets 
have  equal  ratios,  or  when  the  first  is  the  same  part  or  mul- 
tiple of  the  second,  as  the  third  is  of  the  fourth.  Thus,  3, 
6,  4,  8,  and  a,  ar,  6,  br,  are  geometrical  proportionals. 

ar        br 
For  I  =  I  =  2,  and  —  =  —  =  r.     And  they  are  stated  thus, 

a        b 
3  :  6  :  :  4  :  8,  &c. 

Direct  Proportion  is  when  the  same  relation  subsists  be- 
tween the  first  term  and  the  second,  as  between  the  third  and 
the  fourth  :  As  in  the  terms  above.  But  reciprocal,  or  In- 
verse Proportion,  is  when  one  quantity  increases  in  the  same 
proportion,  as  another  diminishes  :  As  in  the'ce,  3,  6,  8,  4  ; 
and  these,  a,  ar,  br,  b. 

The  Quantities  are  in  geometrical  progression,  or  con- 
tinuous proportion,  when  every  two  terms  have  always  the 
same  ratio,  or  when  the  first  has  the  same  ratio  to  the  second, 
as  the  second  to  the  third,  and  the  third  to  the  fourth,  &c. 
Thus,  2,  4,  8,  16,  32,  64,  &c.  and  a,  ar,  ar^ ,  ar^,ar'^,  ar^ , 
&c.  are  series  in  geometrical  progression. 

The  most  useful  part  of  geometrical  proportion  is  contained 
in  the  following  theorems  ;  which  are  similar  to  those  in 
Arithmetical  Proportion,  using  multiplication  for  addition,  &c. 

1.  When 


GEOMETRICAL  PROPORTION.  229 

1.  When  four  quantities  are  in  geometrical  proportion,  the 
product  of  the  two  extremes  is  equal  to  the  product  of  the 
two  means.  As  in  these,  3,  6,  4,  8,  where  3X8  =  6X4  = 
24  ;  and  in  these,  a,  ur,  6,  6r,  where  a  X  6r  =  ar  X  6  = 
abr. 

2.  When  four  quantities  are  in  geometrical  proportion, 
the  product  of  the  means  divided  by  either  of  the  extremes 
gives  the  other  extreme.  Thus,  if  3  :  6  :  :  4  :  8,  then 
6X4  6X4 

=  8,  and  =  3  ;  also  if  a  :  ar  i  :  b  :  br,  then 

3  8 
abr                     abr 
=  6r,  or =  a.     And  this  is  the  foundation  of  the 

a  br 

Rule  of  Three. 

3.  If,  any  continued  geometrical  progression,  the  product 
of  the  two  extremes,  and  that  of  any  other  two  terms,  equally 
^listant  from  them,  are  equal  to  each  other,  or  equal  to  the 
square  of  the  middle  term  when  there  is  an  odd  number 
of  them.  So  in  the  series  1,  2,  4,  8,  16,  32,  64,  &c.  it  is  1 
X  64  =  2  X  32  =  4  X  16  =  8  X  8  =  64. 

4.  In  any  continued  geometrical  series,  the  last  term  is 
equal  to  the  first  multiplied  by  such  a  power  of  the  ratio  as 
is  denoted  by  1  less  than  the  number  of  terms.  Thus,  in  the 
series  3,  6,  12,  24,  48,  96,  4-c.  it  is  3  X  ^^  =  96. 

6.  The  sum   of  any  series  in   geometrical  progression,  is 
found  by  multiplying  the  last  term  by  the  ratio,  and  dividing 
the  difference  of  this   product  and  the  first  term  by  the  dif- 
ference between  1  and  the  ratio.     Thus,  the  sum  of  3,  6, 
192  X  2-3 

12,  24,  48,  96,  192,  is =  384-3  =  381.     And 

2—1 
the  sum  of  n  terms  of  the  series,  a,  ar,  ar^^  ar^,  ar'^,  &c.  to 
ar"~^  X  r  —  a       ar^  —  a       r"  —  1 

ar»~^t  is  — — •  = = a. 

r— 1  r— 1  r  —  1 

6.  When  four  quantities,  o,  «r,  6,  br,  or  2,  6,  4,   12,  are 
proportional  ;  then  any  of  the  following  forms  of  those  quan 
tities  are  also  proportionl,  viz. 

1.  Directly,      a    :  ar  :  :  b    :  6r  ;  or  2  :  6  :  ;     4  :  12. 

2.  Inversely,    ar  :  a   :  :  br  :  b    ;  or  6  :  2  :  :  12  :     4. 

3.  Alternately,  a    :  b    ::  ar  :  br  ;  or  2  :  4  :  :    6:12. 

4.  Com- 


S.MO  ALGEBRA. 

4.  Compoundedly,  a  :  a-\-ar  :  :b  :  b-\-br  ;  or  2  :  8  :  :  4  :  16. 

5.  Dividedly,  a  ;  ar  — a  :  :  6  :  6r  — 6  ;  or  2  :  4  :  ;  4  :  8. 

e.  Mixed,ar4-a  :  ar  —  a  :  :  6r  +6  :  6r— 6  ;  or  8  :  4  :  :  16  :  8. 
7.  Multiplication,  ac  :  arc  :  :  be  :  6rc  ;  or  2.3  :  6.3  :  :  4  :  12. 

a     ar 
S.   Division,  —  :  —  :  :  fc  :  fer ;  or  1  :  3  :  :  4  :  12. 

c      c 
9.  The  numbers  a,  6,  c,  d,  are  in  harnaonical  proportion, 
when  a  :  d  :  :  a  ^  b  :  cm  d  \  or   when  their  reciprocal? 
1111 

— , — ,  — ,  — ,  are  in  arithmetical  proportion. 
abed 


EXAMPLES. 

1.  Given  the  first  term  of  a  geometric  series  1,  the  ratio 
2,  and  the  number  of  terms  12  ;  to  find  the  sum  of  the  series  ? 
First,  1  X  2»  1  =  1  X  2048,  is  the  last  terra. 

2048  X  2  —  1        4096  -  1 

Then = =  4095,  the  sum  required. 

2  ~  1  1 

2.  Given  the  first  term  of  a  geometrical  series  i,  the  ratio 
i,  and  the  number  of  terms  8  ;  to  find  the  sum  of  the  series  ? 
First,  i  X  {\Y  =  1  X  yig  =  3^^,  is  the  last  term. 
Then,a~^i^  X  1)  ^(1-  i)  =  (i-^x-)4.^  =f|f  X  ^ 

=  Iff.  the  sum  required. 

3.  Required  the  sum  of  12  terms  of  the  series  1,  3,  9,  '27, 
31,  &c.  Ans.  266720. 

4.  Required  the  sum  of  12  terms  of  the  series  1,  \,\,  ^V^ 
3V,  &c.  Ans.  fff IH. 

5.  Required  the  sum  of  100  terms  of  the  series  J,  2,  4,  8, 
16,  32,  &c.  Ans.  1267650600228229401496703205376. 

See  more  of  Geometrical  Proportion  in  the  Arithmetic, 


SIMPLE  EQUATIONS. 

An  Equation  in  the  expression  of  two  equal  quantities,  with 
the  sign  of  equality  (=)  placed  between  them.  Thus,  10— 
4  =  6  is  an  equation,  denoting  the  equality  of  the  quantities 
10-4  and  6. 

Equations 


SIMPLE  EQUATIONS.  23) 

Equations  are  either  simple  or  compound.*  A  Simple 
Equation,  is  that  which  contains  only  one  power  of  the  un- 
known quantity,  without  including  different  powers.  Thus, 
j;  —  a  =  6  -f"  c>  or  ax^  =  6,  is  a  simple  equation  containing 
only  one  power  of  the  unknown  quantity  x.  But  x^  —  2a3' 
aaai  6^  li  a  compouud  one. 

GENERAL  RULE. 

Reduction  of  Equations,  is  the  finding  the  value  of  the 
unknown  quantity.  And  this  consists  in  disengaging  that  quan- 
tity from  the  known  ones  ;  or  in  ordering  the  equation  so,  that 
the  unknown  letter  or  quantity  may  standalone  on  one  side  of 
the  equation,  or  of  the  mark^  of  equality,  without  a  co-effi- 
cient ;  and  all  the  rest,  or  the  known  quantities,  on  the  other 
side.-T-In  general,  the  unknown  quantity  is  disengaged  from 
the  known  ones,  by  performing  always  the  reverse  operations. 
So  if  the  known  quantities  are  connected  with  it  by  -{-or  ad- 
dition, they  must  be  subtracted  ;  if  by  minus  (  -),  or  subtrac- 
tion, they  must  be  added  ;  if  by  multiplication,  we  must  divide 
by  them  ;  if  by  division,  we  must  multiply  ;  when  it  is  in  any 
power,  we  must  extract  the  root  ;  and  when  in  any  radical, 
we  must  raise  it  to  the  power.  As  in  the  following  particular 
rules  ;  which  are  founded  on  the  general  principle  of  perform- 
ing  equal  operations  on  equal  quantities  ;  in  which  case  it 
is  evident  that  the  results  must  still  be  equal,  whether  by 
equal  additions,  or  subtractions,  or  multiplications,  or  divisions, 
or  roots,  or  powers. 

PARTICULAR  RULE  L 

When  known  quantities  are  connected  with  the  unknown 
by  ^  or  —  ;  transpose  them  to  the  other  side  of  the  equation, 
and  change  their  signs.  Which  is  only  adding  or  subtracting 
the  same  quantities  on  both  sides,  in  order  to  get  all  the  un- 
known terms  on  one  side  of  the  equation,  and  all  the  known 
ones  on  the  other  side*. 

Thus, 


*  Here  it  is  earnestly  recommended  that  the  pupil  be  accustom-, 
ed,  at  every  line  or  step  »n  the  reduction  of  the  equations,  to 
name  the  paiticiilar  operation  to  be  performed  on  the  equation 
in  the  Ime,  in  order  to  pr<'duce  the  next  form  or  state  of  the  equa- 
tion, tn  applying  each  of  these  rules,  according  as  the  particular 
fornos  of  the  equatipn  may  require  ;  applying  them  according  to  the 

order 


332  ALGEBRA. 

Thus,  if  a:-j-5  =  ^  »  then  transposing  5  gives  x  =  8 — 3  =  3. 
And,  ifa;'-3-{-7=9;  then  transposing  the  3,  and  7,  gives 

x—9  +  3-7=6. 
Also,    '\i  x—a  -\-  b  =  cd  :  then  by   transposing  a   and  h, 

it  is  a:  =  a  —  6  -|-  cd. 
In  like  manner,  if  6x  —  6  =  4a:  4- 10,  then  by  transposing 
6  and  4a:,  it  is  6a;  —  4j:  =  10  -|-  6,  or  a;  =  IQ. 

RULE  II. 

When  the  unknown  term  is  multiplied  by  any  quantity-  ; 
divide  all  the  terms  of  the  equation  by  it. 

Thus,  if  ao;  =  ab  —  4a  ;  then  dividing  by  a,  gives  a;=6  —  4. 
And,   if  3:c  -|-  6  =  20  ;  then  first  transposing  5  gives  2>x 
=  16  ;  and  then  by  dividing  by  ^,  it  is  x  =6. 
In  like  manner,  if  ax-\-^ab  =  4c^  ;  then  by  dividing  by  a,  it 
4c2  4c2 

is  x-{-3b  = ;  and  thfen  transposing 36,  gives  x= — '■ — 3b, 

a  a 

RULE  m. 

When  the  unknown  term  is  divided  by  any  quantity  ;  we 
must  then  multiply  all  the  terms  of  the  equation  by  that  divi- 
sor ;  which  takes  it  away. 

X 

Thus,  if-  =  34-2  :  then  mult,  by  4,  gives  x  =  12+8=20. 
4 

X 

And,  if  -  =  36  +  2c  ^  d  : 
a 
then  by  mult,  a,  it  gives  x  =  Sab  +  2ac  —  ad. 
3x 

Also,  if 3  =  5  +  2; 

6 
Then  by  transposing  3,  it  is  |j7  =  10. 
And  multiplying  by  6,  it  is  3x  =  60. 
Lastly  dividing  by  3  gives       x  =  16f . 


•rder  in  which  they  are  here  placed  ;  and  beginning  every  Lne  with 
the  words  Then  iy,  as  in  the  following  specimens  of  Examples; 
which  two  words  will  always  bring -to  his  recollection,  tht,t  be  is  to 
pronuuncf-  what  particular  operation  he  is  to  perform  on  the  last  line, 
in  order  to  gi  ve  the  next  j  allotting  always  a  single  line  for  each  ope- 
ration, and  ranging  the  equations  neat'y  just  under  each  other,  in  the 
several  lines,  as  they  are  successively  produced. 

RULE 


SIMPLE  EQUATIONS.  233 

RUU5  IV. 

Whew  the  unknown  quantity  is  included  in  any  root  op 
surd  ;  transpose  the  rest  of  the  terms,  if  there  be  any,  by 
Rule  I  ;  then  raise  earh  side  to  such  a  power  as  is  denoted  by 
the  index  of  the  surd  ;  viz.  square  each  side  when  it  is  the 
square  root  ;  cube  each  side  when  it  is  the  cube  root  ;  &c. 
which  clears  that  radical. 

Thus,  if  ^j;— 3  =  4  ;  then  transposing  3,  gives  ^x  =  7  ; 

And  squaring  both  sides  gives  x  =  49. 


And,  if  y/2x  -1-10  =  8: 

Then  by  squaring,  it  becomes  2x  -}-  \0  =  64  ; 
And  by  transposing  10,  it  is  So:  =  54  ; 
Lastly,  dividing  by  2,  gives  x  =  27. 


Also,  if  1/  3x-f4  -{-3  =  6: 


Then  by  transposing  3,  it  is  ^  3a:  -{-  4  =:  3  ; 
And  by  cubing,  it  is  3ar  +  4  =  27  ; 
Also,  by  transposing  4,  it  is  3v  =  23  ; 
Lastly,  dividing  by  3,  gives  a:  =  7|. 

RULE  V. 

When  that  side  of  the  equation  which  contains  the  un* 
knovvn  quantity  is  a  complete  power,  or  can  easily  be  reduced 
to  one,  by  rule  1,  2,  or  3  :  then  extract  the  foot  of  the  said 
power  on  both  sides  of  the  equation  ;  that  is,  extract  the 
square  root  when  it  is  a  square  power,  or  the  cube  root  when 
it  is  a  cube,  &c. 

Thus,  if  a;2  -(-  8a:  +  16  =  36,  or  (x  -f-  4)2  =  36  : 

Then  by  extracting  the  roots,  it  is  x  -j-  4  =  6  ; 

And  by  transposing  4,  it  is  a;  s=  6  —  4  =  2. 
Andif3r2_19  =  21-h35. 

Then,  by  transposing  19,  it  is  3x^  =  75  ; 

And  dividing  by  3,  gives  x'^  =25  ; 

And  extracting  the  root,  gives  a;  =  5. 
Also,  if5a;2— 6  =  24. 

Then  transposing  6,  gives  f  a:^  =  30  ; 

And  multiplying  by  4,  gives  3x-  =  120  ; 

Then  dividing  by  3,  gives  a:^  =  40  ; 

Lastly,  extracting  the  root,  gives  a;  =  -^Z  40  =:  6-324555. 

Vol.  I.  31 

RULE 


S34  ALGEBRA. 


RULE  VI. 


When  there  is  any  analogy   or  proportion,  it  is   te  be 
changed  into  an  equation,   by  multiplying  the  two  extreme 
terms  together,  and  the  two  means  together,  and  making  the 
One  product  equal  to  the  other. 
Thus,  if  2j:  :  9  :  :  3  :  5. 
Then,  mult,  the  extremes  and  means,  gives  lOx  =  27  ; 
And  dividing  by  10,  gives  «.==  2^^. 

And  if  ^x  :  a  :  :  5b  :  2c. 
Then  mult,  extremes  and  means  gives  fca:  =  bab  ; 
And  multiplying  by  2,  gives  3cx  =  lOab  ; 

lOab 

Lastly,  dividing  by  3c,  gives  x  = . 

3c 
Also,  if  10— a;  :  fa;  :  :  3  :  1. 
Then  mult,  extremes  and  means,  gives  10— a:  =  2x  j 
And  transposing  x,  gives  10  =  3.r  ; 
Lastly,  dividing  by  3,  gives  3^  =  x. 

RULE  vn. 

When  the  same  quantity  is  found  on  both  sides  of  art 
equation,  with  the  same  sign,  either  plus  or  minus,  it  may  be 
left  out  of  both  :  and  when  every  term  in  an  equation  is  either 
multiplied  or  divided  by  the  same  quantity  it  may  be  struck 
out  of  them  all. 

Thus,  if  3a:  4-  2a  =  2a  -f  6  : 

Then  by  taking  away  2a,  it  is  3a;  =  5. 

And,  dividing  by  3,  it  is  a;  =  ^b. 
Also  if  there  be  4aa;  -f  6ab  —  lac. 

Then  striking  out  or  dividing  by  a,  gives  4.r  4'  66  =  Ic. 

Then,  by  transposing  66,  it  becomes  4j7  =  7c  —  66  ; 

And  then  dividing  by  4  gives  a;  =  |c  —  |6. 

Again,  if  fa;  -  |  =  V  —  h 
Then,  taking  away  the  |,  it  becomes  fa;  =  y  ; 
And  taking  away  the  3's,  it  is  2jr  =  10  ; 
Lastly,  dividing  by  2  gives  j:  =  5. 

MISCELLANEOUS  EXAMPLES. 

1.  Given  7a;  -  18  =  4a:  -f-  6  ;  to  find  the  value  of  x. 
First,  transposing  18  and  5x  gives  3a;  =  24  ; 
Then  dividing  by  3,  gives  x  ==  8. 

2.  Givcm 


SIMPLE  EQUATIONS.  235 

2.  Given  ?0  —  4j?  —  12  =  92—  lOx  ;  to  find  x. 
First  transposing  20  and  12  and  lUa:,  gives  6j?  =  84  ; 
Then  dividing  by  6,  gives  x  =  14. 

3.  Let  4«x— 56  =  3rfx  -f-  2c  be  given  ;  to  find  x. 

First,  by  trans.  66  and  3(ix,  it  is  4ax  —  3dar  =  56  +  2c  : 

56  4-  2c 

Then  dividing  by  4a  •—  3c?,  gives  x  = — ^. 

4a  —  Ji 

4.  Let  5jr2_l2jr:  =  9j;  -{-  2^2  be  given  ;  to  find  x. 
First,  by  dividing  by  j?,  it  is  5-c  — 12  =  9  +  2-c  j 
Then  transposing  12  and  2x,  gives  3j:  =  21  ; 
Lastly,  dividing  by  3,  gives  x  =  7. 

5.  Given  9ax^  —  15a6a:2  =  6ax^  -{-  12a:c2  .  to  find  x. 
First,  dividing  by  3ax^ ,  gives  3jc  — 56  =  2a;  -h  4  ; 
Then  transposing  56  and  2a;,  gives  x  =-  Sb  -{-  4, 


^.  Let 1 =  2  be  given,  to  find  x. 

3         4         5 
First,  multiplying  by  3,  gives  r  —  ^x  -j-  fa:  =  6  , 
Then  multiplying  by  4,  gives  x  +  */a;  =  24. 
Also  multiplying  by  5,  gives  11  x  =  120. 
Lastly,  dividing  by  17,  gives  x  =  lj\, 

x—B       X  X— 10 

7.  Given -] =12 ;  to  find  x. 

3  2  3 

First,  mult,  by  3,  gives  jr-5  -f  fjr  =  36  —  :r  +  10 
Then  traasposing  5  and  x,  gives  2x  -f-  f  ^  =  61  ; 
And  multiplying  by  2,  gives  7a;  =  102. 
Lastly,  dividing  by  7,  gives  x  =  14|. 

3a; 

8.  Let  ^ h  7  =  10,  be  given  ;  to  find  x. 

First,  transposing  7,  gives  y^f  a?  =  3  ;  r 

Then  squaring  the  equation,  gives  f  x  =  9  ; 
Then  dividing  by  3,  gives  ^x  =  3  ; 
Lastly,  multiplying  by  4,  gives  x  =  12. 

6a= 


9.  Let  2x  4-  2  V«^-f -ra  =  :;/"-2'q:^i-'  be  given  ;  to  findx. 

First,  mult,  by  ^a^  -{-a:- ,  gives  2x  ^  a^  -{-x^  ■i-2a2  -\-2x^ 
=  5a2. 

Then  trans.  2a3  and  2x^,  gives  2x  y/a^  -i-ar^  =  Sa^  -.2a:«  ; 

Then 


Vse 


ALGEBRA. 


Then  by  squaring,  it  is  4*^  X  a^+x-  =  3a2  ^^a-*^*  ; 

That  is,  4a2j^2   ^  ^j^a  =  9a*  ~   VZa^x^  +  4a-*   ; 

By  taking  4x*  from  both  sides,  itis  402^^2=9^4  ^  i2a^x^; 

Then  transposing  12a2a;3,  gives  16a2a;2,  =  9a*  ; 

Dividing  by  a^ ,  gives  16a:2  =  9a2  ; 

And  dividing  by  16,  gives  x^  =  J^a^  ; 

Lastly  extracting  the  root,  gives  a:  =  |a. 

EXAMPLES  FOR  PRACTICE. 

1.  Given  2x  —  6-4-16  =  21  ;  to  find  or.      Ans.  jr  =  6 

2.  Given  9a;  —  15  =  x  4-  6  :  to  find  x,      Ans.  x  =  2|* 

3.  Given  8— 3a;-|-l2=30  — 6a:-f  4  ;  to  find  x.  Ans.  a;  =  7, 

4.  Given  x  +  j-^— 4^  =  13  ;  to  find  x.         Ans.  x  =  12. 

5.  Given  3a:  +  \x-\-  2  =  5jc— 4  ;  to  find  x.     Ans.  a:  =  4. 

6.  Given  4aa7  +  ^a  —  2  =  aa;  —  bx  ;  to  find  ar. 

6— a 

Ans.  X  = 

Sa+36 

7.  Given  ^x-^^x  -{-  la?  =  i  ;  to  find  x.      Ans.  a?  =  f^. 

8.  Given  ^  4  -f  a:  ==  4  — ^a: ;  to  find  x.     Ans.  ar  =  2i, 

a72 

9.  Given  4a  -{-,x  = ;  to  deter,  x,    Ans  x  =--2a, 

4a+ar 


10.  Given  y/4a^  +  ar2  =  */  46*  +  x^  j  to  find  a;. 

6*~-4a* 

Ans.  X  z=  ^ « 

2a2 

4a__ 

11.  Given  ^  x  -{-  ^  2a -{- x  =^2^   •  ^  ;  to  find  x. 

Ans.  X  =  |a. 
a  a 

12.  Given { =  26  ;   to  find  a;. 

1  +2a:        1  —  2a: 

6-« 

Ans.  x=i^ . 

6 


13.  Given  o  -f-  a:  =  ^a^-\-x>^  46a  ^  ^.a  .  to  findx. 

Ans.  X  =  ^ «. 

«  a 

OF 


SIMPLE  EQU^TIONTS.  J37 

©F  REDUCING  DOUBLE,  1 RIPLF.  &c.  EQUATIONS,  COVTAININ© 
TWO,  THREE,  OR  MO^tE  UNKNOWN  QUANTITIES. 

PROBLEM  I. 

To  Exterminate  Two  Unknown  Qnantities  ;  Or^  to  Reduce  the 
Th}0  Simple  Equations  containing  the/n,  to  a  Single  one, 

RULE  L 

Find  the  value  of  one  of  the  unknovvn  letters,  in  terms  of 
the  other  quantities,  in  each  of  the  equations,  hy  the  methods 
already  explained.  Then  put  those  two  values  equal  to  each 
other  for  a  new  equatiorr,  with  only  one  unknown  quantity  in 
it,  whose  value  is  to  be  found  as  before. 

JVote.  It  is  evident  that  we  must  first  begin  to  find  the 
values  of  that  letter  which  are  easiest  to  be  found  in  the  two 
proposed  equations. 

EXAMPLES. 

1.  Given    j5^J:2y=uS'  *"^°^^^"^^- 

In  the  1st  equat.  transp.  3y  and  div.  by  2,  gives  x  = ; 

2 
14  +  22/ 

In  the  2d  transp.  2y  and  div.  by  5,  gives  x  = ; 

5 
14  +  23^       ll-Sy 

Putting  tlifese  two  values  eaual,  gives = ; 

5  2 

Then  mtilt.  by  6  and  2,  gives  28  -{-  4i/  =  85  —  15^/ ; 
Transposing  28  and  loy,  gives  19t/  =  &7  ; 
And  dividing  by  19,  gives  y  =  3. 
And  hence  x  =  4. 
Or,  to  do  the  s^me  by  finding  two  values  of  y,  thus  : 

17-2a: 

In  the  1st  equat.  tr.  2a;  and  div.  by  3,  gives  y  = ; 

3 
Sr— 14 

In  the  2d  tr.  Zy  and  14,  and  div.  by  2,  gives  y  = —  ; 

2 
6ar— 14       17 -2x 

Putting  these  two  values  equal,  gives = ; 

2  3 

Mult,  by  2  and  by  3,  gives  16ar-42  =  34-4x  ; 

Transp. 


2Sa  ALGEBRA. 

Transp.  42  and  4a:,  gives  lOar  =  76  ; 

Dividing  by  19,  gives  a:  =  4.  '      . 

Hence  y  =  3,  as  before. 

2.  Given  JpH^r^J;  to  find  a;  and  j/. 

Ans.  x  =  a  -}-  b,  and  y  =  la  -^  i j*. 

3.  Given  3x  -\-  y  =^  22,  and  3i/  -{-  a;  =  18  ;  to  find  x  and  y. 

Ans.  X  =  6i  and  y  =  4. 

4.  Given  5  f    T  i    H  q,  J  ;  to  find  ^  and  y. 

i    3"^  "T   32/  —  y 3    > 

Ans.  J?  =  6,  and  y  =  3, 
2a7       31/       22         3jc       22/       67 

k.  Given {-  — .  =  — ,  and f*  —  =  —  ;  to  find  x  andy. 

3         5         5  5         3        15 

Ans.  a;  =  3,  and  y  =  4. 

6.  Given  or  +  ^y  =  s,  and  jt^  —  41^2  =  c/2  .  ^q  find  a:  and  y. 

s2  -I-  t^a  s3  _  d^ 

Ans.  J?  = ,  and  y  =■ . 

25  4s 

7.  Givea^F  —  2y  =  d,  and  x  :  t/  :  :  a  :  6  ;  to  find  x  and  y. 

ad  bd 

Ans.  :r  = ,  and  y  = . 

a  —  26  a  —  26 


RULE  n. 

Find  the  value  of  one  of  the  unknown  letters,  in  only  one 
ef  the  equations,  as  in  the  former  rule  ;  and  substitute  this 
value  instead  of  that  unknown  quantity  in  the  other  equation, 
and  there  will  arise  a  new  equation,  with  only  one  unknown 
quantity,  whose  value  is  to  be  found  as  before. 

jsfote.  It  is  evident  that  it  is  best  to  begin  first  with  that 
letter  whose  value  is  easiest  found  in  the  given  equations. 

EXAMPLES.  ^ 

1.  Given  I  ?^  _  1^  Z  \l  I  ;  to  find  x  and  y. 

This  will  admit  of  four  ways  of  solution  ;  thus  :  First, 

17-32/ 

In  the  1st  eq.  trans.  3y  and  div.  by  2,  gives  x  == ; 

2 
85—162/ 
This  val.  subs,  for  x  in  the  2d,  gives, ■  2y=14  ; 

Mult,  by  2,  this  becomes  8&  —  16y  — .  41/  =  28  j 

Transp. 


SIMPLE  EQUATIONS.  23? 

Transp.  I6y  and  4y  and  28,  gives  67  =  1%  ; 
And  dividing  by  19,  gives  3  =  t/. 
.17— 3y 

Then  x  == =  4. 

2 

14-f2y 

2dly,  in  the  2d  trans.  2y  and  div.  by  6,  gives  x  = —  ; 

5 
28-f4^ 

This  subst.  for  x  in  the  1st,  gives {-  3y  =  17  ; 

6 
Mult,  by  6,  gives  28  +  4y  -f  ISy  =  85  j 
Transpos.  28.  gives  19y  =  57  ; 
And  dividing  by  1 9,  gives  y  =  3. 
144-21/ 

Then  x  = =  4,  as  before. 

5 

17-2a: 

3dly,  in  the  1st  trans.  2x  and  div.  by  3,  gives  y  = ; 

3 
34— 4a; 

This  subst  for  y  in  the  2d,  gives,  5a? =  14  ; 

3 
Multiplying  by  3  gives     16a:  —  M  -|-  4*  =  42  ; 
Transposing  34,  gives       19a;  =  76  ; 
And  dividing  by  1 9,  gives     a:  =    4. 
17— 2x 

Hence  y  = =  3,  as  before. 

3 

6a;- 14 

4thly,  in  the  2d  tr.  2y  and  14  and  div.  by  2,  gives  y  = .  <• 

2 
16a;  -  42 

This  substituted  in  the  1st,  gives  2x  -| =  17  ; 

2 
Multiplying  by  2,  gives  1 9a;- 42  =  34  ; 
Transposing  42,  gives  19a;  =  76  ; 
And  dividing  by  19,  gives  x  =  4. 
oa:-  14 

Hence  y  = =  3,  as  before. 

2 

2.  Given  2a;  +  3y  =  29,  and  3jr  -  2^/  =  11  ;  to  find  x  and  y, 

Ans.  a?  =  7,  and  y  =^  5. 

3.  Given  l^ly^  '2^;  tofinda;andy. 

Ans.  X  =  8,  and  y  =  6. 
4.  Given 


240  ALGEBRA. 

4.  Given  }  ^^  ^  '  '  L_'  oq  J  J  ^o  find  x  and  y. 

Ans.  X  =  6,  and  y  =  4. 

X  y 

5.  Given f-  3y  =  21,  and h3x  =  29  ;  to  find  x  and  y, 

3  3 

Ans.  a;  =±  9,  and  y  =  6. 

•^  y  ^  — y         X 

6.  Given  10 =  ^  ^  4,  and 1 2  = 

2  3  2  4 
32/— x 
■--!  ;  to  find  X  and  y.                   Ans.  x=  S,  dmdy  =  6. 


7.  Given  x  :  y  :  :  4  :  3,  and  x^  -^  y^  =  SI ;  to  find  x  and  y. 

Ans.  a;  =  4,  and  y  =  3> 


RULE  TIL 

Let  the  given  equations  be  so  multiplied,  or  divided,  &c. 
and  by  such  numbers  or  quantities,  as  will  make  the  terms 
which  contain  one  of  the  unknown  quantities  the  same  in 
both  equations  ;  if  they  are  not  the  sanie  when  first  pro- 
posed. 

Then  by  adding  or  subtracting  the  equations,  according  as 
the  sines  may  require,  there  will  remain  a  new  equation,  with 
only  one  unknown  quantity,  as  before.  That  is,  add  the  two 
equations,  when  the  sines  are  unlike,  but  subtract  them  when 
the  signs  are  alike,  to  cancel  that  common  term. 

JV()ie.  To  make  two  unequal  terms  become  equal,  as  above, 
multiply  each  term  by  the  co-efficient  of  the  other. 


EXAMPLES. 

«-«°S2^  +  5  =  16?"°''''^"'°"'^- 
Here  we  may  either  make  the  two  first  terms,  containing 
•c,  equal,  or  the  two  2d  terms,  containing  y,  equal.  To  make 
the  two  first  terms  equal,  we  must  multiply  the  1st  equation 
by  2,  and  the  2d  by  5  ;  but  to  make  the  two  2d  terms  equel, 
we  must  muhiply  the  1st  equation  by  5,  and  the  2d  by  3  ;  as 
follows* 

l.By, 


SIMPLE  EQUATIONS.  241 

1.  By  making  the  two  first  terms  equal  : 

Mult,  the  1st  equ.  Wy  2,  gives         lOx  —    6y  =  18 
And  mult  the  2d  by  5,  gives  IOjt  -f-  2by  ="80 

Subtr.  the  upper  from  tHe   under,  gives  Sly  =  62 
And  dividing  by  31,  gives  y  =■    ^. 

9  +  3y 

Hence,  from  the  1st  given  equ.  x  = =    3. 

5 

2.  By  making  the  two  2d  terms  equal : 
Mult,  the  1st  equat.  by  5,  gives  2bx —  {by  =  4b  j 
And  mult,  the  2d  by  3,  gives  6x  -{■  15r/  =  48  ; 
Adding  these  two,  gives                Six  =  93  ; 
And  dividing  by  31 ,  gives  x  =    S  ; 

bx  —  9 
Hence,  from  the  1st  equ.  y  = =  2. 


MISCELLANEOUS  EXAMPLES. 

^4-8  y-\-6 

1.  Given |-%=21,and — '■ h  Bar  =  23  ;  to  find  x 

4  3 

and  y,  Ans.  x  =  4,  and  y  =  S. 

Sx  —  y  Sy  -i-  X 

2.  Given h  10=  13,  and \~  b  ==  12;  to  find 

4  2 

X  and  y.  Ans.  a:  =  5,  and  y  =  S. 

Sx  -{'  4y       X  6x  —  2?/        y 

3.  Given 1 =  It),  and -|-_-.=  i4;to 

5  4  3  6 

find  X  and  y.  Ans  :c  =  8,  and  y  =  4. 

4.  Given  Sir  +  4y  =  38,  and  4x  —  Sy  =  9  ;  to  find  x  and  y. 

Ans.  X  =  6,  and  y  =  b, 

PROBLEM  XL 

To  Exterminate  Three  or  More  Unknown  Quantities  ;  Or,  to 
Reduce  the  Simple  Equations,  containing  them,  to  a  Single 
one. 

RULE. 

This  may  be  done  by  any  of  the  three  methods  in  the  last 
problem  :  viz. 

1.  After  the  manner  of  the  first  rule  in  the  last  problem, 
find  the  value  of  one  of  the  unknown  letters  in  each  of  the 
given  equations  :  next  put  two  of  these  values  equal  to  each 
other,  and  then  one  of  these  and  a  third  value  equal,  and  so 
on  for  all  the  values  of  it ;  which  gives  a  new  set  of  equations, 

Vol.  1.  32  with 


242  ALGEBRA. 

with  which  the  same  process  is  to  be  repeated,  and  so  oa  til! 
there  is.onl}^  one  equation,  to  he  reduced  by  the  rules  for  a 
single  equation. 

2.  Or,  as  in  the  2d  rule  of  the  same  problem,  find  the  value 
of  one  of  the  unknown  quantities  in  one  of  the  equations 
only  ;  then  substitute  this  value  instead  of  it  in  the  other 
equations  ;  which  gives  a  new  set  of  equations  to  be  resolved 
as  before,  by  repeating  the  operation. 

3.  Or,  as  in  the  3d  rule,  reduce  the  equations,  by  multiply- 
ing or  dividing  them,  so  as  to  make  some  of  the  terms  to  agree  : 
then,  by  adding  or  subtracting  them,  as  the  signs  may  require, 
one  of  the  letters  may  be  exterminated,  &c.  as  before. 


EXAMPLES. 

2/4-    ^=    9) 
1.  Given  <  x  -}-  2y  +  32-  =  16  >  ;  to  find  x,  y,  and  z. 
■-3^4-4^  =  31^ 


1.  By  the  1st  method  : 
Transp.  the  terms  containing  t/  and  z  in  each  equa.  gives 
X  =■    9 —    y —    ^, 
jr=  16  — 2y  —  3z, 
a7  =  2I  —  3y— 42-; 
Then  putting  the  1st  and  2d  values  equal,  and  the  2d  and  34 
values  equal,  give 

9—    2/—    2r=  16  — 2y— 3z, 
16  —  22/  —  32  =  21  -  3^  —  42  ; 
lu  the  I  St  trans.  9,  z,  and  2y,  gives  y  =  7  —  2^  ; 
1b  the  2d  trans.  16,  32  and  3y,  gives  y  =  6  —     z  \ 
Putting  these  two  equal,  gives  3  —  2r  =  7  —  2^  ; 
Trans.  5  and  22,  gives  z  =«:  2. 
Hence  ?/=  6  —  2  =  3,  and  37  =  9  —  ^ —  ^  =  4. 

'      2dly.  By  the  2d  method  : 

From  the  1st  equa.  ^  =  9  —y  —  ^  \ 
This  value  of  x  substit.  in  the  2d  and  3d,  gives 
9  -f    y  -f-  2z  =  16, 
9  -f  2y  4-  3^  =  21  ; 
In  the  1st  trans.  9  and  22,  gives  y  =  7  —  22  ; 
This  substit.  in  the  last,  gives  23  —  2-  =  21  j 
Trans,  z  and  21,  gives  2  =  2. 
Hence  again  y  =  7  —  22  =  3,  and  jc  =  9  —  y  —  2r  =  4. 

3dly.  By 


SIMPLE  EqUATIONfi.  243 

3dly.     By  the  3d  method  ;  subtracting  the   1st  equ.  from 
the  2d,  and  the  2d  from  the  3d,  gives 

y  -f  2^  =  7, 
y  -\-    2  =  5; 
Subtr.  the  latter  from  the  former,  gives  2  =  2. 
Hence  y  =  6  —  2r  .=  3,  and  re  =  9 -— y  —  z  =  4. 


{    X  -h3y  -{-  2^  =  38)    ;  to 

i  ^  +  ^y  +  i^=  1^') 

Ans.  X 

i    ^  +  i3/  +  i^=27) 
Given  {    :r-hJ-2/  + 1^  =  20}   5 
(    ^  +  i2/  +  i^=  J^) 


2.  Given  <    jp  -j-  3y  4-  2^  =  38  )   ;  to  find  ar,  y,  and  2. 
10  5 
Ans.  X  =  4f  y  =  6,  2  = 


to  tind  x,  y,  and  ^. 
Ans.  i:  =  1,^  =  20,  2"  =  60. 

4.  Given  x  —  y  =  2,  x  —  2  =  3,  and  y  —  2  =  1;  to 
find  or,  y,  and  z.  Ans.  J7  =  7;  y  =  6;  2:=, 4. 

^  2^  -{-  3t/  4-  4^  =  34  ^ 

5.  Given  ?  3x  -j-   >?/  -f  62-  =  46  J  ;  to  find  a:,  y,  and  z. 

(  4x  +  5y  4- 6^  =  68  ) 


A  COLLECTION  OF  QUESTIONS  PRODUCING  SIMPLE 
EQUATIONS. 


Quest.  1.  To  find  two  nambers,  such,  that  their  sum  shall 
be  10,  and  their  diflference  .6. 

Let  X  denote  the  greater  number,  and  y  the  less*. 
Then,  by  the  1st  condition  j:  -f-    y  —  10, 
And  by  the  2d         -         -    x  ^—    y  =    6^ 
Transp.  y  in  each,  gives      a:  =  10  —    y, 
and  jr  =    6  -f-    y  ; 
Put  these  two  values  equal,  gives  6  -{-    y  =     10  —  y  ; 
Transpos.  6  and  —  y,  gives    -    2y  =  4  ; 
Dividing  by  2,  gives       -  -      y  ==  2. 

And  hence  -         -         -     x  =x.  6  -h  y  =  ^' 


*  In  all  these  solutions,  as  many  unknown  letters  are  always  used 
as  there  are  unknown  numbers  to  be  found,  purposely  the  better  to 
exercise  the  modes  of  reducing  the  equations  :  avoidinj^  the  short 
ways  of  notation,  which  though  giving  a  shorter  solution,  are  for  that 
reason  less  useful  to  the  pupi!.  as  affording  less  exercise  in  practising 
the  several  rules  in  reducing  equations. 

Quest.  2. 


244  ALGEBRA. 

Quest.  2.     Divide    100/.   among  a,  b,  c,  so  that  a   may 
have  20/.  more  than  b,  and  b  10/.  more  than  c.    , 

Let  X  =  a's  share,  y  =  b's,  and  c  =  c's. 
Then  :x:  -^  y  -\-    z  =  100, 
a:  =  2/  4-  ^0, 
>y  ^  2  ~\-  10. 
In  the  1st  substit.  y  -{-  20  for  :r,  gives  2^/  -f  z  +  20  =  100  ; 
In  this  substituting  z  -\-  10  for  y,  gives  3z  -f-  40  =  100  y 
By  transposing  40,  gives     -        '-       '  32  =*  GO  ; 
And  dividing  by  3,  gives      -         -  z  =  20 

Hence  y  =  z  -\-  10  =  30,  and  :c  =  ^  +  20  =  60. 

Qbest.  3.    A  prize  of  600/.  is  to  be  divided  between  two 
persons,  so   as  their  shares  may  be  in  proportion  as  7  to  8  ;  < 
required  the  share  of  each. 

Put  X  and  y  for  the  two  shares  ;  then  by  the  question, 

t  :  S  :  :  j^  :  y,  or  mult,  the  extremes 
and  the  means,  7j/  =  8x, 
^  and  X  -\-  y  =  600  ; 

Transposing  y,  gives  x  =■  500  —  y  ; 
This  substituted  in  the  1st,  gives  ly  ="4000—  Sy  ; 
By  transposiujg  By,  it  h  lay  =  4000  ; 
By  dividing  by  15,  it  gives  y  =  266|  ; 
And  hence  j:  =  600  —  j/  =  233i. 

Quest.  4.  What  number  is  that  whose  4th  part  e^iceeds 
its  6th  part  by  10  ? 

Let  X  denote  the  number  sought. 
Then  by  the  question  ^x  —  J-o:  =  10  ; 
By  mult,  by  4,  it  becomes  x  —  ijc  =  40  ; 
By  mult,  by  6  it  gives  x  =  200,"  the  number  sought. 

Quest.  5.  What  fraction  is  that  to  the  numerator  of 
which  if  1  be  added,  the  value  will  be  |  ;  but  if  one  be  added 
to  the  denominator,  its  value  will  be  J- ? 

x 
Let  —  denote  the  fraction. 

y 

.       X  -{-'  1  -^x 

Then  by  the  quest. =  ^,  and =  i. 

y        ^  2/  +  1 

The  1st  mult,  by  2  and  y,  gives  2x  -\-  2  =^  y  ; 
The  2d  mult,  by  3  and  ?/  -f  1  is  3j7  ==  y  +  1  ; 
The  upper  taken  from  the  under  leaves  x  —  2=1; 
By  trans pos.  2,  it  gives  x  =  3. 
And  hence  2/  =  2a:  -|-  2  =  8  ;  and  the  fraction  is  |. 

Quest.  6., 


SIMPLE  EQUATIONS.  245 

QuBST.  6.  A  labourer  engaged  to  serve  for  30  days  on 
'those  conditions :  that  for  every  day  he  worked,  he  was  to 
receive  20d^  but  for  every  day  he  played,  or  was  absent,  he 
was  to  forfeit  lOd.  Now  at  the  end  of  the  ime  he  had  to 
receive  just  20  shillings,  or  240  pence.  It  is  required  to  find 
how  many  days  he  worked,  and  how  many  he  was  idle  ? 

Let  jc  be  the  days  worked,  and  y  the  days  idle. 
Then  20x  is  the  pence  earned,  and  lOy  the  forfeits  ; 
Hence,  by  the  question    -    x  -\^  y  =  30y 

andSOx—  101/  =  240; 
The  1st  mult,  by  10,  gives  Wx  -f  Wy  =  300  ; 
>  These  two  added  give  -  30x  =  640  ; 
This  div.  by  30,  gives  -       or  =  18,  the  days  worked  ; 
Hence         -         2/ =  30  —  x  =  12,  the  days  idled. 

Quest.  7.  Out  of  a  cask  of  wine,  which  had  leaked  away  i, 
30  gallons  tvere  drawn  ;  and  then,  being  gaged,  it  appeared 
to  be  halt  full  ;  how  much  did  it  hold  ? 

Let  it  be  supposed  to  have  held  x  gallons. 
Then  it  would  have  leaked  ^x  gallons, 
Conseq.  there  had  been  taken  away  ^x  -\-  30  gallons. 
Hence  ir  =  ix  +  30  by  the  question. 
,     Then  mult,  by  4,  gives  2x  =  :¥  -{-  120  ; 
And  transposing  x,  gives  x  =  120  the  contents. 

Quest.  8.  To  divide  20  into  two  such  parts,  that  3  times 
the  one  part  added  to  5  times  the  other  may  make  76. 

Let  X  and  y  denote  the  two  parts. 
Then  by  the  question  ^  -  j7  -f-  ^  =  20, 
and  3x  4-  52/  =  76. 
Mult,  the  1st  by  3,  gives  -  -  3x  -j-  3t/  =  60  ; 
Subtr.  the  latter  from  the  former,  gives  2y  =  16  ; 
And  dividing  by  2,  gives     -         -         -  y  z=    Q. 

Hence,  from  the  1st,  -         x=20  —  y  =  12. 

Quest.  9.  A  market  woman  bought  in  a  certain  number  of 
eggs  at  2  a  penny,  and  as  many  more  at  3  a. penny,  and  sold 
them  all  out  again  at  the  rate  of  5  for  two-pence,  and  by  so 
doing;  contrary  to  expectation,  found  she  lost  3d. ;  what  num- 
ber of  eggs  had  she  ? 

Let  X  =  number  of  eggs  of  each  sort. 
Then  will  |x  =  cost  of  the  first  sort, 
And  ^x  =  cost  of  the  second  sort  j 

But 


246  ALGEBRA. 

But  5  :  2  :  :  2jr  (the  whole  numher  of  eggs)  :  ^x ; 
Hence  f  x  =  price  of  both  sorts,  at  6  for  2  pence  ; 
Then  by  the  question  ^x  +  i-^  —  f^  —  3  ; 
Mult,  by  2,  gives     -    x  +  ^x  —  |a7  =  6  ; 
And  mult,  by  3, gives  6.r  —  2_4^  =  jg  . 
Also  mult,  by  5,  gives  x  =  90,  the  number  of  egg»  of 
each  sort. 

Quest.  10.  Tvro  persons,  a  and  b,  engage  at  play.  Before 
they  begin,  a  has  80  guineas,  and  b  has  60.  After  a  certain 
number  of  games  won  and  lost  between  them,  a  rises  with 
three  times  as  many  guineas  as  b.  Query,  bow  many  guineas 
did  a  win  of  b  ? 

Let  X  denote  the  number  of  guineas  a  won. 
Then  a  rises  with  80  -{-  x. 
And  B  rises  with  60  —  x  ;  / 
Theref  ,by  the  quest.  80  +  ^  =  180  —  3a:  ; 
Transp.  80  and  3jr,  gives  4jr  =  100  ; 
And  dividing   by   4,  gives    x  =    25,  the  guineas  won. 

QUESTIONS  FOR  PRACTICE. 

1.  To  determine  two  numbers  such,  that  their  difference 
may  be  4,  and  the  difference  of  their  squares  64. 

Ans.  6  and  10. 

2.  To  find  two  numbers  with  these  conditions,  viz.  that 
half  the  first  with  a  3d  part  of  the  second  may  make  9,  and 
that  a  4th  part  of  the  first  with  a  fifth  part  of  the  second  may 
make  5.  Ans.  8  and  15. 

3.  To  divide  the  number  20  into  two  such  parts,  that  a  3d 
of  the  one  part  added  to  a  fifth  of  the  other,  may  make  6. 

Ans.  15  and  5. 

4.  To  find  three  nurnbers  such,  that  the  sum  of  the  1st  and 
2d  shall  be  7,  the  sum  of  the  1st  and  3d  8,  and  the  sum  of  the 
2d  and  3d  9.  Ans.  3,  4,  6. 

6.  A  father,  dying,  bequeathed  his  fortune,  which  was 
2800L  to  his  son  and  daughter,  in  this  manner  ;  that  for  every 
half  crown  the  son  might  have,  the  daughter  was  to  have  a 
shiUing.     What  then  were  their  two. shares  ? 

Ans.  The  son  2000/.  and  the  daughter  800/ 

6.  Three  persons,  a,  b,  c,  make  a  joint  contribution, 
which  in  the  whole  amounts  t©  400Z.  :  of  which  sum  b  con- 

tributeg 


SIMPLE  EQjUATIONS.  247 

tributes  twice  as  much  as  a  and  20/.  more  ;  and  c  as  much  as 
k  and  B  together.     What  sum  did  each  contribute  ? 

Ans.  A  60/.  B  140/.  and  c  200/. 

7.  A  person  paid  a  bill  of  100/,  with  half  guineas  and  crowns, 
using  in  all  202  pieces  ;  how  many  pieces  were  there  of  each 
sort  ?  Ans.  180  half  guineas,  and  22  crowns. 

8.  Says  a  to  b,  if  you  give  me  10  guineas  of  your  money, 
I  shall  then  have  twic^  as  much  as  you  will  have  left ;  but 
says  B  to  A,  give  me  10  of  your  guineas,  and  then  I  shall  have  3 
times  as  many  as  you.     How  many  had  each  ? 

Ans.  A  22,  B  26. 

9.  A  person  goes  to  a  tavern  with  a  certain  quantity  of 
money  in  his  pocket,  wh*^re  he  spends  2  shillings  ;  he  then 
borrows  as  much  money  as  he  had  left,  and  going  to  another 
tavern,  he  there  spend  2  shillings  also  ;  then  borrowing  again 
as  much  money  as  was  left,  he  went  to  a  third  tavern, 
where  likewise  he  spent  2  shillings  ;  and  thus  repeating  the 
same  at  a  fourth  tavern,  he  then  had  nothing  remaining.  What 
sum  had  he  at  ^rst  ?  Ans.  3s.  9d, 

10.  A  man  with  his  wife  and  child  dine  together  at  an  inn. 
The  landlord  charged  1  shilling  for  the  child  ;  and  for  the 
woman  he  charged  as  much  as  for  the  child  and  i  as  much  as 
for  the  man  ;  and  for  the  man  he  charged  as  much  as  for  the 
woman  and  child  together.     How  much  was  that  for  each  ? 

Ans.  The  woman  20d.  and  the  man  32c/. 

11.  A  cask,  which  held  60  gallons,  was  filled  with  a  mix- 
ture of  brandy,  wine,  and  cyder,  in  this  manner,  viz.  the 
cyder  was  6  gallons  more  than  the  brandy,  and  the  wine  was 
as  much  as  the  cyder  and  }  of  the  brandy.  How  much  was 
there  of  each.  Ans.  Brandy  15,  cyder  21,  wine  24. 

12.  A  general,  disposing  his  army  into  a  square  form,  finds 
that  he  has  284  men  more  than  a  perfect  square  ;  but  increas- 
ing the  side  by  1  man,  he  then  wants  25  men  to  be  a  complete 
square.     Then  how  many  men  had  he  under  his  command  ? 

Ans.  24000. 

13.  What  number  is  that,  to  which  if  3,  5,  and  8,  be 
severally  added,  the  three  sums  shall  be  in  geometrical  pro- 
gression ?  Ans.   1. 

14  The  stock  of  three  traders  amounted  to  860/.  the 
shares  of  the  first  and  second  exceeded  that  of  the  third 

by 


248  ALGEBRA. 

by  240  ;  and  the  sum  of  the  Sfd  and  3d  exceeded  the  6rst  by 
360.     What  was  the  share  of  each  ? 

Ans.  The  1st  200,  the  2d  300,  the  3d  260. 

15.  What  two  numbers  are  those,  which,  being  iij  the  ratio 
of  3  to  4,  their  product  is  equal  to  12  times  thojraum  ? 

Ans.  21  and  28. 

16.  A  certain  company  at  a  tavern,  when  they  came  to 
settle  their  reckoning,  found  that  had  there  beeil  4  more  in 
company,  they  might  have  paid  a  shilling  a-piece  less  than 
they  did  ;  but  that  if  there  had  been  3  fewer  in  company,  they 
must  have  paid  a  shilling  a-piece  more  than  they  did.  What 
thenvvas  the  number  of  persons  in  company,  what  each  paid, 
and  what  was  the  whole  reckoning  ? 

Ans.  24  persons,  each  paid  Is,  and  the  whole 
reckoning  8  guineas. 

17.  A  jocky  has  two  horses  :  and  also  two  saddles,  the  one 
yalued  at  18/  the  other  at  3/.  Now  when  he  sets  the  better 
saddle  on  the  1st  horse,  and  the  worse  on  the  2d,  it  makes  the 
first  horse  worth  double  the  2d  :  but  when  he  places  the  bet- 
ter saddle  on  the  2d  horse,  and  the  worse  on  the  first,  it  makes 
the  2d  horse  worth  three  times  the  1st.  What  then  were 
the  values  of  the  two  horses  ?    Ans.  The  Ist  61.  and  the  2di9Z. 

18.  What  two  numbers  are  as  2  to  3,  to  each  of  which  if  6 
te  added,  the  sums  will  be  as  4  to  5  ?  Ans.  6  and  9. 

19.  What  are  those  two  numbers,  of  which  the  greater  is 
to  the  less  as  their  sum  is  to  20,  and  as  their  difference  is  to 
10  ?  Ans.  15  and  45. 

20.  What  two  numbers  are  those,  whose  difference,  sum, 
and  product,  are  to  each  other,  as  the  three  numbers  2,  3,  5  ? 

Ans.  2  and  10. 

21.  To  find  three  numbers  in  arithmetical  progression,  of 
which  the  first  is  to  the  third  as  5  to  9,  and  the  sum  of  all  three 
is  63  ?  Ans.    16,  21,  27. 

22.  It  is  required  to  divide  the  number  24  into  two  such  parts, 
that  the  quotient  of  the  greater  part  divided  by  the  less,  may 
be  to  the  quotient  of  the  less  part  divided  by  the  greater,  as  4 
to  1.  Ans.  16  and  8. 

23.  A  gentleman  being  asked  the  age  of  his  two  sons, 
answered,  that  if  to  the  sum  of  their  ages  18  be  added, 
the  result  will  be  double  the  age  of  the  elder :  but  if  6  be 

taken 


Q,UADRATIC  EQUATIONS.  249 

taken  from  the  difference  of  their  ages,  the  remainder  will 
be  equal  to  the  age  of  the  younger.  What  then  were  their 
ages  ?  Ans.  30  and  12. 

24.  To  find  four  numhers  such,  that  the  sum  of  the  1st, 
2d,  and  3d,  shall  be  13  ;  the  sum  of  the  1st,  2d,  and  4th,  15  ; 
the  sum  of  the  1st,  3d,  and  4th,  1&  ;  and  lastly  the  sum  of  the 
2d,  3d,  and  4th,  20.  Ans.  2,  4,  7,  9. 

26.  To  divide  48  into  4  such  parts,  that  the  1st  increased 
by  3,  the  second  diminished  by  3,  the  third  multiplied  by  3, 
^nd  the  4th  divided  by  3,  may  be  all  equal  to  each  other. 

Ans.  6,  12,  3,  27. 


QUADRATIC  EQUATIONS. 

Quadratic  Equations  are  either  simple  or  compound. 

A  simple  quadratic  equation,  is  that  which  involves  the 
square  of  the  unknown  quantity  only.  As  ax^  =  b.  And 
the  solution  of  such  quadratics  has  been  already  given  in 
simple  equations. 

A  compound  quadratic  equation,  is  that  which  contains  the 
square  of  the  unknown  quantity  in  one  t^rm,  and  the  first 
power  in  another  term.     As  ax^  -\-  bx  =  c. 

All  compound  quadratic  equations,  after  being  properly 
reduced,  fall  under  the  three  following  forms,  to  which  they 
must  always  be  reduced  by  preparing  them  for  solution. 

1.  x^  -i-  ax  ==  b 

2.  ^2  —  rtJT  =   6 

3.  x^ — ax  =  —  b. 

The  general  method  of  solving  quadratic  equations,  is  by 
what  is  called  completing  the  square,  which  is  as  follows  : 

1.  Reduce  the  proposed  equation  to  a  proper  simple  form, 
as  usual,  such  as  the  forms  above  ;  namely,  by  transposing 
all  the  terms  which  contain  the  unknown  quantity  to  one 
side  of  the  equation,  and  the  known  terms  to  the  other  ; 
placing  the  square  term  first,  and  the  single  power  second  ; 
dividing  the  equation  by  the  co-efficient  of  the  square  or 
first  term,  if  it  has  one,  and  changing  the  signs  of  all  the 
terms,  when  that  term  happens  to  be  negative,  as  that  term 
must  always  be  made  positive  before  the  solution.  Then 
the  proper  solution  is  by  completiag  the  square  as  follows, 
viz. 

VdL.  I.  s^i  2.  Complete» 


280  ALGEBRA. 

2*  Complete  the  unknown  side  to  a  square,  in  this  man- 
ner, viz.  Take  half  the  co  elhcient  of  the  second  term,  and 
square  it  ;  which  square  add  io  both  sides  of  the  equation, 
then  that  side  which  contains  the  unknown  quantity  will  be  a 
complete  square. 

3.  Then  extract  the  square  root  on  both  sides  of  the 
equation*,  and  the  value  of  the  unknown  quantity  will  be 

determined, 


*  As  the  square  root  of  any  quantity  may  be  either  +  or  — ,  there- 
fore all  quadratic  equations  admit  of  two  solutions.  Thus,  the  square 
root  of  4-  «*  is  either  -^  n  or  —  n  ;  fov  +  n  X  +  «  and  —  n  X  -"  " 
are  each  equal  to  -f.  n*.  But  the  square  root  of  —  n*,  or  ^Z  -"  ^*» 
is  imaginary  or  impossible,  as  neither  4*  n  nor  —  n,  when  squared, 
gives  -   n2. 

^  So,  in  the  first  form,  x2  +  aj7  =  ^,  where  a:  •{'  la  is  found  «=  y/ 
h  +  la^,  the  root  may  be  either  +  ^6-f-  ^*,  or  -*-  ^^  +  ia«, 
since  either  of  them  being  multiplied  by  itself  produces  b  +  fa^- 
And  this  ambiguity  is  expressed  by  writing  the  uncertain  or  double 
sign  dt  before  ^yb  +  ^a*  ;  thus  or  =»  :t  ^b  -f  la^  —  ^a- 


In  this  form,  where  a:  =«  ±  ■•^  +  k'l^  —  ia.  the  first  value  of 
X,  viz.  X  ■=»  +  y/b  +  |:02  —  ^a.  in  always  affirmative  ;  for  sincSe 
ifl2  -j-  b  is  greater  than  la 2,  the  greater  square  must  neces- 
sarily have  the  greater  root  ;  therefore  ^  b  -h  ^a^  will  always 
be  greater  than  ^ja^,  or  its  equal  ia ;  and  consequently  -f- 
^b  +  \aS   —  |a  will  always  be  affirmative. 

The  second  value,  viz.  07  *=  —  <^b  +  |a*  —  ^a  will  always  be 
negative,  because  it  is  composed  of  two  negative  terms.  Therefore 
when  x^  -^  ax  ==*  b,  we  shall  have  j:  ==  -f-  y/b  -J-  ^a^  — .  |a  for  the 
affirmative  value  of  x,  and  or  =*  -f-  <yb  +■  Ja*  —  ^a  for  the  nega- 
tive value  of  X. 


In  ths  second  form,  where  x  =-  ±:.s/b  -|-  ia*  +  ^a  the  first 
value,  viz.  a:  -«  +  ^^4-  i«*  +  ia  is  always  affirmative,  since  it 
is  composed  of  two  affirmative  terms.  But  the  second  value,  viz* 
a:  =="  —  v^  b  +  Ja2  -f-  |a,  will  always  be  negative ;  for  since 
b  -{-  ia^  is  greater  than  ^a^,  therefore  ^b  -j  l'.'^  will  be  greater 
than  v/:iu2,  or  its  equal  \a  ;  and  consequently  --  y/b  +  \a^  +  ^a  is 
always  a  negative  quantity. 

Therefore, 


QUADRATIC  EQUATIONS.  251 

determined,  making  the  root  of  the  known  side  either  -f  or 
— ,  which  will  give  two  roots  of  the  equation,  or  two  values 
•f  the  unknown  quantity. 

KotCy  1.  The  root  of  the  first  side  of  the  equation,  is 
always  equal  to  the  root  of  the  first  term,  with  half  the 
co-efficient  of  the  second  term  joined  to  it,  with  its  sign, 
whether  -f-  or  — . 

2.  All  equations,  in  which  there  are  two  terms  including 
the  unknown  quantity,  and  which  have  the  index  of  the  one 
just  double  that  of  the  other,  are  resolved  like  quadratics, 
by  completing  the  square,  as  above. 

Thus,  J?*  -f  ax2  =^  6,  or  jr^n  _|,  ^^.n  _..  ^^  ^^  ^^  ^^.2  s=j^ 
are  the  same  as  quadratics,  aad  the  value  of  the  unknown 
quantity  may  be  determined  accordingly. 


Therefore,  when  x^  -ar  =  6,  we  shall  have  a:  =  4  V'^  +  i«^ 
+  \a  for  the  affiimative  value  of  x  ;  and  x  =  -  y/b  -f  ia^  -f-  \a 
for  tbe  negative  value  of  ar  ;  so  that  in  both  the  first  and  secnd 
forms,  the  unknown  quantity  has  always  two  values,  one  of  which  is 
positive,  and  the  other  nogative. 

But  in  the  third  form,  where  x  —  ±.  ^ ia^  —  6  +  ^a,  both  the 
values  of  x  will  be  posHive  when  |o2  is  greater  than  b.  For  ihe 
first  value,  viz.  x  =^  •\-  »/  \a^  -  6-f"i^  will  then  be  aflSrmative, 
being  composed  of  two  affirmative  terms. 

The  second  value,  viz.  x  =■  —  v'  i'^^  —  6  -J-  |a  is  affiima- 
tive also  J  for  since  \a^  is  greater  than  \a^  -  6,  thecefo-e  »/  ,\a^  or 
^  is  greater  than  ^  \a'^  -  Aj  and  consequently  —  v^  :Ju*—  6  +  ia 
,will  always  be  an  afii  mative  qjantity.  So  that»  when  jr*  —  ax 
«=  -.  d,_we  shill  have  x  =  -f-  ^  }a2  —  A  +  ^a,  And  also  x  =  — 
*/  ia2  «.A  -f.  ifl,  for  the  vaUes  of  Xt  both  positive. 

But  m  this  third  form,  if  6  be  greater  .than  ^a*,  the  solution  of 
the  proposed  question  will  be  impossible.  For  since  the  square  of 
any  quantity  (whether  that  quantity  be  affirmative  or  negative; 
is  always  affirmative,  the  square  root  of  a  negative  quantity  is  im- 
possible, and  cannot  be  assigned.  But  when  h  is  greater  than 
^a*,  then  ^,r,2— A  is  a  negative  quantity;  and  therefore  its  root 
^  i'Z«  — ^  is  impossible,  or  imaginary  ;  consequently,  in  that  case, 
^  =  i^  ±  >/  i'J^— ^»  oi*  tbe  two  roots  or  values  of  x^  are  both  im  - 

possible,  or  imaginary  qwantities. 

EXAMPLES. 


252  ALGEBRA. 

EXAMPLES. 

1.  Given  x^  -|-  4jr  =  60 ;  to  find  x. 

First,  by  completing  the  square,  x^  -f-  4^  +  4  =  64  ,' 

Then  by  extracting  the  root,  j?  +  2  =  ±  8  ; 

Then,  transpos  2,  gives,  j:  =  6  or  —  10,  the  two  roots. 

2.  Given  x^  —  6ar4-10  =  66  ;  to  find  x. 
First  trans.  10  gives  x^  —  6a:  =  65  ; 

Then  by  complet.  the  sq.  it  is  x^  —  6j7  -f  9  =  64  ; 
And  by  extr.  the  root,  gives  j?— 3  =  ±:  8  ; 
Then  trans.  3,  gives  x  =  1 1  or  -  6. 

3.  Given  2x2  -f  8x  -  30  =  60  ;  to  find  x. 
First  by  transpos,  20,  it  is  2x3  -f  8x  =  90  ; 
Then  div.  by  2,  gives  x^  -}"  4.r  =  45  ; 

And  by  compl.  the  sq.  it  is  x^  +  4x  -|-  4  =  49  ; 
Then  extr.  the  root,  it  is  x  +2  =  ±  7  ; 
And  transp.  2,  gives  x  ==  5  or  — -  9. 

4.  Given  3x2  -,  3x  +  9  =  ^  y  to  find  x. 
First  div.  by  3,  gives  x*  —  x  -}-  3  =  2f  ; 
Then  transpos.  3,  gives  x2  — .  x  =  —  f  ; 
And  compl.  the  sq.  gives  x2  —  x  -\-  \  =■  -^  ] 
Then  extr.  the  root  gives  x  —  ^  =  ±  |  ; 
And  transp.  i,  gives  x  =  |  or  ^. 

5.  Given  ^x^  —  ^x  +  30^  =  52|,   to  find  x. 
First  by  transpos  30^,  it  is  ^x^  —  \x  =  22i  ; 
Then  mult,  by  2  gives  x^  —  |x  =  44^  ; 

And  by  compl.  the  sq.  itis  x2  —  2.J7  -j-  i  =  44A  ■ 
Then  extr.  the  root,  gives  x  —  i  =  ^^  6|  j 
And  transp.  ^,  gives  x  =  7  or  —  6^. 

6.  Given  ax^  —^bx  =  c;  to  find  x. 

b  c 

First  by  div.  by  a,  it  is  x2 x  =  —  ; 

a  a 

b  b  c         b^ 

Then  compl.  the  sq.  gives~x2 x  -j = ! ; 

a  4a2       a       4a' 

b  4ac  +  fc3 

And  extrac.  the  root,  gives  x =  ±  ^ —  • 

2a  4a2 

b  4ac  +  i2         6 

Then  transp.  — ,  gives  x  =  ±  y/ 1 . 

2o  4a2  2a 

7.  Given  x*  -.2ax2  =  t ;  to  find  x. 

First  by  compl.  the  aq.  gives  x*  —  Sax*  ^  a^  =  a2  -j-  ^  ; 

And 


QUADRATIC  EQUATIONS.  263 


And  extract,  the   root,  gives  z^  —  a  —  ±.  ^ofi  •\-  h ; 

Then  transpos.  a,  gives  x^  =  ±  ^a^  -^  b  -{-  a  ; 

And  extract,  the  root,  gives  x=±:^a±^a'-\-b. 
And  thns.  by  always  using  similar  words  at  each  line,  the 
pupil  will  resolve  the  following  examples. 

EXAMPLES  FOR  PRACTICJB. 

1.  Given  a:2,r-  6x  —  7  =  33  ;  to  find  x,  Ans.  x  =  10. 

2.  Given  x^  -, —  5x  —  10  =  14  ;  to  find  a;.  Ans.  a;  =  6. 

3.  Given  5x^  -f  4a;  —  90  =  114 ;  to  find  x.  Ans.  a:  =  6* 

4.  Given  |a;2  — ix  -{-  2  =  9  ;  to  find  a:.  Ans.  a:  =  4. 
6.  Given  3a:*  — 2x*  =  40  ;  to  find  x.  Ans.  a;  =  2. 

6.  Given  ^x  —  i^  a:  =  1^^  ;  to  find  x.  Ans.  ac  =  9. 

7.  Given  ^x^  -f  f  a;  =  f  ;  to  find  x.        Ans.  x  =  -727766, 

8.  Given  x^  -\-  4x^  =  12 ;  to  find  x, 

Ans.  a:  =  3/  2  =  1-269921. 

^.  Given  a;^  +  4a;  =  a^  -f  2  ;  to  find  x, 

Ans.  X  =  v^ oMhe— 2. 

QUESTIONS  PR0DUCIN6  QUADRATIC  EQUATIONS. 

1.  To  find  two  numbers  whose  difference  is  2,  and  product  80. 
Let  X  and  y  denote  the  two  required  numbers*. 

Then  the  first  condition  gives  x  —  y  =  2, 

And  the  second  gives  xy  =  80. 

Then  transp.  y  in  the  1st  gives  x  =  y  -{-  2  ; 

This  value  of  x  substitut.  in  the  2d,  is  y^  -f-  2i/  =  80 

Then  comp.  the  square  gives  y-  -]-  2y  -{-  I  =^  Ql  ; 

And  extrac.  the  root  gives  y  -{-  1  =  9  ; 

And  transpos.  1  gives  y  =  8  ; 

And  therefore  a;  =  y  4*  2  =  10. 


♦  These  questions*  like  those  in  siniple  equations)  aire  also  solved 
by  using  as  tnany  unknown  letters,  as  are  the  numbers  required,  for 
^e  better  exercise  in  reducing  equations  ;  not  aiming  at  the  shortest 
modes  of  solution,  which  would  not  affoi"d  so  much  useful  practice. 

2.  To 


264  ALGEBRA. 

2.  To  divide  the  number  14  into  two  such  psirta,  that  their 
product  may  be  48. 

Let  X  and  y  denote  the  two  numbers. 

Then  the  1st  condition  gives  x  -{-  y  =  14, 

And  the  2d  gives  xy  =  4S. 

Then  transp^  y  in  the  1st  gives  x  =  14  —  y ; 

This  value  subst.  for  x  in  the  2d,  is  14y  —  y^  =  43  j 

Changing  all  the  signs,  to  make  the  square  positive, 

gives  y^  —  14y  =  —  48  ; 
Then  compl.  the  square  gives  y^  —  Hy  +  49  =  1  ; 
And  extrac.  the  root  gives  y  —  7  =  ±  1  ; 
Then  transpos.  7,  gives  ?/  =  8  or  6,  the  two  parts. 

3.  Given  the  sum  of  two  numbers  =  9,  and  the  sum  of 
their  squares  =  46  ;  to  find  those  numbers. 

Let  X  and  y  denote  the  two  numbers. 

Then  by  the  1st  condition  or  -{-  2/  =  9. 

And  by  the  2d  x^  +  y^  =  45. 

Then  transpos.  y  in  the  1  st  gives  jt  =  9  —  1/  » 

This  value  subst  in  the  2d  gives  81  —  '\8y  -{-  2y^  =  45  ; 

Then  transpos.  81,  gives  2y^  —  I82/  =  —  36  ; 

And  dividing  by  2  gives  y^  —  %  =  —  18  ; 

Then  compl.  the  sq  gives  y^  •—  9y  4-  Y  =  f  ; 

And  extrac.  the  root  gives  y  —  f  =  ±  |  j 

Then  transpos.  f  gives  1/  =  6  or  3,  the  two  numbers. 

4.  What  two  numbers  are  those,  whose  sum,  product,  and 
difference  of  their  squares,  are  all  equal  to  each  other  ? 

Let  X  and  y  denote  the  two  numbers. 
Then  the  1st  and  2d  expression  give  x  -\-  y  =  xy^ 
And  the  1st  and  3d  give  x  -\-  y  =■  x^  — y^. 
Then  the  last  equa.  div.  hy  x  -\-  y^  gives  1  =  jr-  —  y. ; 
And  transpos.  7/,  gives  y  -{-  \  =^  x  ; 
This  val.  substit-  in  the  1st  gives  2?/  4"  1  =  2/^  +  'iji 
And  transpos.  2y,  gives  1  =  2/^  —  y  ; 
Then  complet.  the  sq.  gives  f  =  2/^  —  2/  +  4  ;. 
And  extracting  the  root  gives  ^  a/  5  =  y  —  ^y 
And  transposing  ^  gives  i  \/  6  -f-  ^  =  ^ ; 
And  therefore  x  =  ^4-l=iv'6  +  t« 
And  if  these  expressions  be  turned  into  numbers,  by  ex- 
tracting the  root  of  6,  &;c.  they  give  a:  =  2  6180  -f-,  and 
2/=  1-6180 -f. 

5.  There  are  four  number^  in  arithmetical  progression,  of 

whioK 


Q.tJADRATIC  EQUATIONS.  26^ 

which  the  product  of  the  two  extremes  is  22,  and  that  of  the 
means  40  ;  what  are  the  numbers  ? 

Let  j:  =  the  less  extreme, 

and  y  =  the  common  difference  ; 
Then  x^x  4-  y,  J?  +  2y,  Ir  4*  %>  will  be  the  four  numbers. 
Hence  by  the  1st  condition  x^  +  Sxy  =  22, 
And  by  the  2d  x^  -f-  3xy  +  2y^  =  40. 
Then  subtracting  the  first  from  the  2d  gives  2i/a  =  18  ; 
And  dividing  by  2  gives  2/^=9; 
And  extracting  the  root  gives  i/  =  3. 
Then  substit  3  for  y  in  the  1st,  gives  x^  -{-  9x  =  22  ; 
And  completing  the  square  gives  x^  -}-  9x  -{■  y  =  *  p  ; 
Then  extracting  the  root  gives  ^  4*  f  =  V  ; 
And  transposing  f  gives  x  =  2  the  least  number. 
Hence  the  four  numbers  are  2,  5,  8,  11. 

6.  To  find  3  numbers  in  geometrical  progression,  whose  sum 
shall  be  7,  and  the  sum  of  their  squares  21. 

Let  X,  y,  and  z,  denote  the  three  numbers  sought. 
Then  by  the  1st  condition  xz  =  y^, 
And  by  the  2d  x  +  y  -{-  2r  =  7, 
And  by  the  3d  x^  -{- y^  +  z^  =  21. 
Transposing  y  in  the  2d  gives  x  -{-  z=  7  -^  y  ; 
Sq.  this  equa.  gives  x^  -\-  2xz  -\-  z^  -f-  =  49 —  14y-{-y^; 
Substi.  2i/2  for  2x2",  gives  x^ -\-2y^ -{-z^  =  49—141/4-2/2; 
Subtr.  y^  from  each  side,  leaves  x^  -{-y^  -{-z^  =  49—  14y ; 
Putting  the  two  values  of  x^  +  1^2  4.  2,2  i  ^^__.q__  ... 
equal  to  each  other,  gives  5  ^^ 

Then  transposing  21  and  14i/,  gives  14i/  =  28  ; 
And  dividing  by  14,  gives  y  =  2. 
Thensubstit.  2  for  y  in  the  1st  equa.  gives  xz  =  4, 
And  in  the  4th,  it  gives  x  +  2  =  5  ; 
Transposing  z  in  the  last,  gives  x  =  5  —  z  ; 
This  substit.  in  the  next  above,  gives  5z  —  z^  =  4  j 
Changing  all  the  signs,  gives  z^  —  5z  =  —  4  ; 
Then  completing  the  square,  gives  z^  —  52-  -|-  y  =  a  j 
And  extracting  the  root  gives  2-  —  f  =  -t  f  ; 
Then  transposing  f  gives  z  and  x  =  4  and  1,  the  two 

other  numbers  ; 
So  that  the  three  numbers  are  1,2,  4. 

QUESTIONS  FOR  PRACTICE. 

1.  What  number  is  that  which  added  to  its  square  makes 
42?  Ans.  6. 

2,  T# 


256  ALGEBRA. 

.2.  To  find  two  numbers  such,  that  the  less  miy  be  to  the 
greater  as  the  greater  is  to  12,  and  that  the  sum  of  their  squares 
may  be  45.  Ans.  3  and  6. 

3.  What  two  numbers  are  those,  whose  difference  is  2,  and 
the  difference  of  their  cubes  98  ?  Ans.  3  and  5. 

4.  What  two  numbers  are  those  whose  sum  is  6,  and  the 
sum  of  their  cubes  72  ?  Ans.   2  and  4. 

5.  What  two  numbers  are  those,  whose  product  is  20,  and 
the  difference  of  their  cubes  G 1  ?  Ans.  4  and  5. 

6.  To  divide  the  number  1 1  into  two  such  parts,  that  the 
product  of  their  squares  may  be  784  Ans.  4  and  7, 

7.  To  divide  the  number  5  into  two  such  parts,  that  the  sum 
of  their  alternate  quotients  may  be  4^,  that  is  of  the  two 
quotients  of  each  part  divided  by  the  other^ 

Ans.  1  and  4. 

8.  To  divide  12  into  two  such  parts,  that  their  product  may 
be  equal  to  8  times  their  difference.  .         Ans.  4  and  8. 

9.  To  divide  the  number  10  into  two  such  parts,  that  the 
square  of  4  times  the  less  part,  may  be  112  more  than  the 
square  of  2  times  the  greater.  Ans.  4  and  6. 

10.  To  find  two  numbers  such,  that  the  sum  of  their  squares 
may  be  89,  and  their  sum  multiplied  by  the  greater  may  pro- 
duce 104.  Ans.  5  and  8. 

11.  What  number  is  that,  which  being  divided  by  the 
product  of  its  two  digits,  the  quotient  is  5i  ;  but  when  9  is 
subtracted  from  it,  there  remains  a  number  having  the  same 
digits  inverted  ?  Ans.  32. 

12.  To  divide  20  into  three  parts,  such  that  the  continual 
product  of  all  three  may  be  270,  and  that  the  difference  of  the 
first  and  second  may  be  2  less  than  the  difference  of  the  second 
and  third.  Ans.  5,  6,  9. 

13.  To  find  three  numbers  in  arithmetical  progression,  such 
that  the  sum  of  their  squares  may  be  56,  and  the  sum  arising 
by  adding  together  once  the  first  and  2  times  the  second  and 
3  times  the  third,  may  amount  to  28.  Ans.  2,  4,  6. 

14.  To  divide  the  number  13  into  three  such  parts,  that 
their  squares  may  have  equal  differences,  and  that  the  sum  of 
those  squares  may  be  75.  Ans.  1,  5,  7. 

15.  To  find  three  numbers  having  equal  differences,  so  that 
their  sum  may  be  12,  and  the  sum  of  their  fourth  powers  962. 

Ans.  3,  4,  6. 

16.  To  find  three  numbers  having  equal  differences,  and 
such  that  the  square  of  the  least  added  to  the  product  of  the 
two  greater  may  make  28,  but  the  square  of  the  greatest  add- 
ed to  the  product  of  the  two  less  may  make  44. 

Ans.  2,  4,  6. 

17.  Three 


CUBIC,  &c.  EqUATIOJ^S.  S5^ 

17.  Three  merchants,  a,  b,  c,  on  comparing  their  gains 
find,  that  among  them  all  they  have  gained  1444/.  ;  and  that 
b's  gain  added  to  the  square  root  of  a's  made  920/  ;  but  if 
added  to  the  square  root  of  c's  it  made  912.  What  were 
their  several  gains  ?  Ans.  a  400,  b  900,  c  144. 

18.  To  find  three  numbers  in  arithmetical  progression,  so 
that  the  sum  of  their  squares  shall  be  93  ;  also  if  the  first  be 
multiplied  by  3,  the  second  by  4,  and  the  third  by  5,  the  sum 
of  the  products  may  be  66.  Ans.  2,  5,  8. 

19.  To  find  four  numbers  such,  that  the  first  may  be  to  the 
second  as  the  third  to  the  fourth  ;  and  that  the  first  may  be 
to  the  fourth  as  1  to  5  ;  also  the  second  to  the  third  as  5  to  9  ; 
and  the  sum  of  the  second  and  fourth  may  be  20. 

Ans.  3.  5,  9,  15. 

20.  To  find  two  numbers  such  that  their  product  added  to 
their  sum  may  make  47,  and  their  sum  taken  from  the  sum 
df  their  squares  may  leave  62.  Ans.  5,  and  7. 


RESOLUTION  OF  CUBIC  AND  HIGHER  EQUATIONS. 

A  Cubic  Equation,  or  Equation  of  the  3d  degree  or  power, 
lis  one  that  contains  the  third  power  of  the  unknown  quantity. 
As  x^  —ax^  +  fca:  =  c. 

A  Biquadratic^  or  Double  Quadratic,  is  an  equation  that 
contains  the  4th  Power  of  the  unknown  quantity  : 
As  X*  —ax^  +  bx^  -"CX  =  d. 

An  Equation  of  the  6th  Power  or  Degree,  is  one  that  con- 
tains the  5th  power  of  the  unknown  quantity  : 
As  x^  ^ax*  -h  bx^     cx^  -{-  dx  =  e. 

And  so  on,  for  all  other  higher  powers.  Where  it  is  to 
be  noted,  however,  that  all  the  powers,  or  terms  in  the 
equation,  are  supposed  to  be  freed  from  surds  or  fractional 
exponents. 

There  are  many  particular  and  prolix  rules  usually  given 
for  the  solution  of  some  of  the  above-mentioned  powers 
or  equations.  But  they  may  be  all  readily  solved  by  the 
following  easy  rule  of  Double  Position,  sometimes  called 
Trial-and-error. 

RULE. 

1.  Find,  by  trial,   two  numbers,  as  near  the  true  toot  as 

^ou  can,  and  substitute  them  separately  in  the  given  equation, 

instead  of  the  unknown  c^uantity  ;  and  find  how  much  the 

V  OL.  i.  ^34  terras 


'< 


26«  ALGEBRA. 

terms  collected  together,  according  to  their  signs  ^^4"  or  — , 
differ  from  the  absolute  known  term  of  the  equation,  marking 
whether  these  errors  are  in  excess  or  defect. 

2.  Multiply  the  difference  of  the  two  numbers,  found  or 
taken  by  trial,  by  either  of  the  errors,  and  divide  the  pro- 
duct by  the  difference  of  the  errors,  when  they  are  alike, 
but  by  their  sum  when  they  are  unlike.  Or  say.  As  the 
difference  or  sum  of  the  errors,  is  to  the  difference  of  the 
two  numbers,  so  is  either  error  to  the  correction  of  its  sup- 
posed number. 

3.  Add  the  quotient,  last  found,  to  the  number  belonging  to 
that  error,  when  its  supposed  number  is  too  little,  but  subtract 
it  when  too  great,  and  the  result  will  give  the  true  root 
nearly, 

4.  Take  this  root  and  the  nearest  of  the  two  former,  or 
any  other  that  may  be  found  nearer  ;  and,  by  proceeding  in 
like  manner  as  above,  a  root  will  be  had  still  nearer  than 
before.     And  so  on  to  any  degree  of  exactness  required. 

Note  1.  It  is  best  to  employ  always  two  assumed  numbers 
that  shall  differ  from  each  other  only  by  unity  in  the  last 
figure  on  the  right  hand  ;  because  then  the  difference,  or 
multiplier,  is  only  1.  It  is  also  best  to  use  always  the  least 
error  in  the  above  operation. 

JVote  2.  It  will  be  convenient  also  to  begin  with  a  single 
figure  at  first,  trying  several  single  figures  till  there  be  found 
the  two  nearest  the  truth,  the  one  too  little,  and  the  other 
too  great ;  and  in  working  with  them,  find  only  one  more 
figure.  Then  substitute  this  corrected  result  in  the  equation, 
for  the  unknown  letter,  and  if  the  result  prove  too  little, 
substitute  also  the  number  next  greater  for  the  second  sup- 
position ;  but  contrarywise,  if  the  former  prove  too  great, 
then  take  the  next  less  number  for  the  second  supposition  : 
and  in  working  with  the  second  pair  of  errors,  continue  the 
quotient  only  so  far  as  to  have  the  corrected  number  to  four 
places  of  figures.  Then  repeat  the  same  process  again  with 
this  last  corrected  number,  and  the  next  greater  or  less,  as 
the  case  may  require,  carrying  the  third  corrected  number 
to  eight  figures  ;  because  each  new  operation  commonly 
doubles  the  number  of  true  figures.  And  thus  proceed  to 
any  extent  that  may  be  wanted. 


EXAMPLES. 

Ex.  1.     To  find  the  root  of  the  cubic  equation  a?^  4-  x^  -f- 
X  =  100,  or  the  value  of  x  in  it. 

H€re 


CUBIC,  &c.  EQUATIONS. 


269 


Here  it  is  soon  found  that 
X  lies  between  4  and  5.  As- 
sume therefore  these  two  num- 
bers, and  the  operation  will  be 
as  follows  : 
1st  Sup. 

4         -         X 

\^        -         x^ 

64         -         a;3 


84 
100 


-16 


-  sums     - 
but  should  be 

-  errors    - 


+  66 


the  sum  of  which  is  71. 
Then  a«  71  :  1  :  ;  16  : 
Hence  x  =  4*2  nearly. 


Again,  suppose  4-2  and  4-3  ; 
and  repeat  the  work  as  fol- 
lows : 


1st  Sup. 
4-2 

17-64 
74-088 


x^ 

<p3 


2d  Sup. 

4-3 

18-49 

79-607 


96-928 
100 


sums       102-297 
100 


—  4-072      errors      -f  2-297 


the  sum  of  which  is  6-369. 
As  6-369  :  1    :  :   2-297  :  0036 
This  taken  from      -     4-300 


leaves  x  nearly  =  4-264 


Again,  suppose  4*264,  and  4-266,  and  work  as  follows  : 
4-264  -         X  -  4-266 

18-181696         -         -ra  -  18-190225 

77-526762         -         x^         -  77-581310 


99-972448 
100 


100-036635 
100 


—0-027552  -     errors      -  -fO-036535 

the  sum  of  which  is  -064087. 

Then  as  -064087  :  -001  :  :  -027562  :  0-0004299 
To  this  adding         -  4-264 


gives  a:  very  nearly  =  4-2644299 


The  work  of  the  example  above  might  have  been  much 
shortened,  by  the  use  of  The  Table  of  Powers  in  the  Arith- 
metic, which  would  have  given  two  or  three  figures  by  in- 
spection. But  the  example  has  been  worked  out  so  particu- 
larly as  it  is,  the  better  to  show  the  method. 

Ex.  2.  To  find  the  root  of  the  equation  x'^~~\hx^  -|-  63  a: 
=  50,  or  the  value  of  x  in  it. 

Here  it  soon  appears  that  x  is  very  little  above  1, 

Suppose 


260 


ALGEBRA. 


Suppose  therefore  1  '0  and  1*1, 
and  work  as  follows  : 

X    -        M 


63:c 
same 


69-3 
-.18-15 
1-331 


62-481 
60 


—  1      -    errors      -  4-2*481 
3-481  sum  of  the  errors. 
As  3-481  :  1  :  :  -1  :  '03  correct. 
1-00 

Hence  or  "=  1  03  nearly 


Again,  suppose  the  two  num- 
bers 103  and   1-02,  &c.  as 
follows  : 
103     >         a:    -   1»02 

64-89     -     63j:      6?2« 
-16-9136  — 15x2-15-6060 
1-092727       x^     l'06120e 


60-069227  sums  49-?  16208 
60  60 


-f  -069227  errors  —-284792 
•284792 


As 


364019:  -01  ::  -069227: 
•0019666 
This  taken  from        1-03 


leaves  x  nearly  =  1  02804 


Note  3.  Every  equation  has  as  many  roots  as  it  contains 
dimensions,  or  as  there  are  units  in  the  index  of  its  highest 
power.  That  is,  a  simple  equation  has  only  one  value  of  the 
root ;  hut  a  quadratic  equation  has  two  values  or  roots,  a  cubic 
equation  has  three  roots,  a  biquadratic  equation  has  four  roots, 
and  so  on. 

And  when  one  of  the  roots  of  an  equation  has  been  found 
by  approximation,  as  above,  the  rest  may  be  found  as  follows. 
Take,  for  a  dividend,  the  given  equation,  with  the  known 
term  transposed,  with  its  sign  changed,  to  the  unknown  side 
of  the  equation  ;  and  for  a  divisor,  take  x  minus  the  root  just 
found.  Divide  the  said  dividend  by  the  divisor,  and  the  quo- 
tient will  be  the  equation  depressed  a  degree  lower  than  the 
given  one. 

Find  a  root  of  this  new  equation  by  approximation,  as  before, 
or  otherwise,  and  it  will  be  a  second  root  of  the  original  equa- 
tion. Then,  by  means  of  this  root,  depress  the  second  equa- 
tion one  degree  lower,  and  from  thence  find  a  third  root,  and 
so  on,  till  the  equation  be  reduced  to  a  quadratic  ;  then  the 
two  roots  of  this  being  found,  by  the  method  of  completing 
the  square,  they  will  make  up  the  remainder  of  the  roots. 
Thus  in  the  foregomg  equation,  having  found  one '  root  to  be 
1-02804,  connect  it  by  minus  with  x  for  a  divisor,  and  the 
equation  for  a  dividend,  &c.  as  follows  : 

^^  1-02804)  x^  ~  \bx^  +  63:j?  —  60  (x^  —  13-97196x  4- 

48-63627=-0 
Theo 


«UBIC,  &c.  EQJJATIONS.  2^t' 

'Then  the  two  roots  of  this  quadratic  equation,  or  —  —  — 
aja  —  13-97196  a:  =  -  48-63627,  by  completing  the  square, 
are  6-57tti3  and  7-39443,  which  are  also  the  other  two  roots 
«f  the  given  cubic  equation.  So  that  all  the  three  roots  of 
that  equation,  viz.  a;^  — 16  x^  -{-  63x  =  60. 

*^^  l*c^f?t  )and  the  sum  of  all  the  roots  is  found  to  be 

A  ^'tltliw^^  ^^^^S  6^"^^    *®   ^^^  co-efficient  of  the 
and  7-c?t^54J  Vg^  ^^^^  ^^  ^^^  equation,  which  the  sum  of 

.  r  /^/^^w^^  I  the  Toots  alwavs  ought  to  be,  when  they  are 
sum    '5  000001   •..  J        &  J  J 

Note  4.  It  is  also  a  particular  advantage  of  the  foregoing 
rule,  that  it  is  not  necessary  to  prepare  the  equation,  as  for 
«ther  rules,  by  reducing  it  to  the  usual  6nal  form  and  state 
of  equations.  Because  the  rule  may  be  applied  at  once  to  an 
unreduced  equation,  though  it  be  ever  so  much  embarrassed 
by  surd  and  compound  quantities.  As  in  the  following  ex- 
ample. 

Ex.  3.  Let  it  be  required  to  find  the  root  x  of  the  equation 
^  144a:a-(x3  -[-  20)^  +  y/  IdQx^ -(x^-^-  24)2"  =  1 H,  or 
the  value  of  x  in  it. 

By  a  few  trials,  it  is  soon  found  that  the  value  of  x  is  but 
little  above  7.  Suppose  therefore  first  that  a;  is  =  7,  and 
then  X  =  8. 

First,  when  x  =  7.  Second,  when  a:  =  8. 

47-906     -     y/  144^2  _  (j.2  -I-  20)2     -     46-476 

65'384     -     v^  196x2  -  (a;2  4- 24)2     .     69-283 

113-290     -     the  sums  of  these  -  115-759 

114-000     -     the  true  number  -  114-000 


— 0-710     -     the  two  errors                ^          -f  1-759 
-fl-759     -  


As  2-469  :  1  :  :  0-710  :  0-2  nearly  . 
7-0 

Therefore  x  =  7-2  nearly 

Suppose  again  x  =  7-2,  and  then,  because  it  turns  out  too 
great  suppose  x  also  =  7-1,  &c.  as  follows  : 

Supp» 


^62- 

ALGEBRA. 

• 

Supp.  X  =  7-2 

Supp. X  = 

71 

47-990     - 

\/  144x2  _  (jr2  4.  20)3 

-     47-973 

66-402     - 

^  196x2  -  (x2  +  24)2 

the  sums  of  these 
the  true  number 

the  two  errors 

::-123:  -024  the  correcti 
7-100  add 

-     65-904 

ir4-392     - 
114  000     - 

113-877 
114-000 

-fO'392     - 
0-123 

— 0-123 

30, 

As     -515  :'l 

Therefore  x  =  7-124  nearly  the  root  required. 

Noit  5.  The  same  rule  also  among  other  more  difficult 
forms  of  equation,  succeeds  very  well  in  what  are  called 
exponential  ones,  or  those  which  have  an  unknown  quantity  in 
the  exponent  of  the  power  ;  as  in  the  following  example  ; 

Ex.  4.     To  find  the  value  of  x  in  the  exponental  equation 

a;^=  100. 

For  more  easily  resolving  such  kinds  of  equations,  it  is 
convenient  to  take  the  logarithms  of  them,  and  then  compute 
the  terms  by  means  of  a  table  of  logarithms.  Thus,  the  loga- 
rithms of  the  two  sides  of  thie  present  equation  are  x  X  log. 
of  a:  =  2  the  log.  of  100.  Then,  by  a  fewt  rials,  it  is  soon 
perceived  that  the  value  of  x  is  somewhere  between  the  two 
numbers  3  and  4,  and  indeed  nearly  in  the  middle  between 
them,  but  rather  nearer  the  latter  than  the  former.  Taking 
therefore  first  x  =  3-6,  and  then  =  3-6,  and  virorking  with 
the  logarithms,  the  operation  will  be  as  follows  : 


First  Supp.  X  =  3-5, 
Log.  of  3'5  =  0-544068 
then  3-5Xlog.  3-5  =  1-904238 
the  true  number  2-000000 


error,  too  little,  — -096762 
•002689 


Second  Supp.  x  =  3*6. 
Log.  of3-6  =  0-656303 
then  3-6  X  log.  3-6  =  2-002689 
the  true  number  2-00000© 


error,  too  great-I-002689 


-098451  sum  of  the  errors.     Then, 


As  -098451  :  -1  : :  -002689  :  0-00273  the  correction 
taken  from  3-60000 


leaves   -    3-59727  =  x  nearly. 


On 


CUBIC,  &c.  EQUATIONS.  263 

On  trial,  this  is  found  to  be  a  very  small  matter  too  little. 
Take  therefore  again,  x  =  3-69727,  and  next  =  3-59728,  and 
repeat  the  operation  as  follows  : 


First,  Supp.  X  =  3-59727. 
Log.  of  3-69727  is  0-655973 
3-59727  X  log. 

of  3-69727  =  1-9999864 

the  true  number  2-0000000 


error,  too  little,  —0-0000146 
--0-0000047 


Second,  Supp.  x  =  3-59728 
Log.  of  3-69728  is  0-565974 
3-59728  X  log. 

of  3-59728  =  1-9999953 

the  true  number  2-0000000 


error,  too  little,  —0-0000047 


0-0000099  diff.  of  the  errors.     Then, 
As  -0000099  :  -00001  :  :  -0000047  ;  -00000474747*  the  cor. 
added  to         -       3-59728000000 


gives  nearly  the  value  of  re  =  3-59728474747 

Ex.  6.    To  find  the  value  of  a:  in  the  equation  sc »  +  lOa:^ 
-f- 6a:  =  260.  Ans.  a;  =  4-1179867. 

Ex.  6.  To  find  the  value  of  x  in  the  equation  x^  —  Src  =  60. 

Ans.3-S648854. 


^  ♦  The  Author  has  here  followed  the  general  rule  in  finding'  as  many 
additional  figures  as  were  known  before :  viz.  The  6  figurcis  3-59728t 
but  as  the  logarithms  here  used  are  to  6  places  only,  we  f.annot  de- 
pend on  more  than  6  .figures  in  the  answer ;  we  have  no  reason, 
therefore,  to  suppose  any  of  the  figures  in  -000004747'47,  to  be 
coprect. 

The  log.  of  3-59728  to  15  places  is  0-55597  42431  34677 
the  log.  of  3«59727  to  15  places  is  0-55597  30358  47267 
which  logarithms  multiplied  by  their  respective  numbers  gi  ve  the  fol- 
lowing products : 

i'S  5?i6  S}  •»*  '""^  *<•  *^  '"*  «e"^- 

Therefore,  the  errors  are        -        -        -        49746  564\88 

and        -        -        -       148773  37702 

and  the  difference  of  errors        -        -        -        99026  812 14. 

Now  since  only  6  additional  figures  are  to  be  obtained,  We  n  my  omit 
the  last  three  figures  in  these  errors  ;  and  state  thuff :  as  diff.  t  >f  errors 
9902681  :  diff.  of  sup.  1  : :  error  4974656 :  the  correction  SI^IT^^.  t,  which 
united  to  3-59728  gives  us  the  true  value  of  a: «  3-597285023  54-     , 


i 


ISx.  7, 


264  ALGEBRA. 

Ex.  7.  To  find  the  value  of  x  in  the  equation  s:^  -|-  2x^ 

—  23  J?  =  70.  Ans.  x  =  5134o7. 
Ex.  8.    To  find  the  value  of  x  in  the  equation  x^  —  VI  x^ 

4-  h^x  =  350.  Ans.  x  =  14-95407. 

Ex.  8.    To  find  the  value  of  x  in  the  equation  jr*  — Sx^ 

—  75a:  =  10000.  Ans.  x  ~  10-2609. 
Ex.  10.  To  find  the  value  of  x  in  the  equation  2j?'*  —  16a: 

4-  40  ar2  —  30j7  =  —  1.  Ans.  x  =  1-284724. 

Ex.   II.    To  find  the  value  of  x  in  the  equation  x^  ■\-  2a;* 

-f  3j73  4-  4a;2  +  6a:  =  54321.  Ans.  x  —  8-414455. 

Ex.  12.    To  find  the  value  of  x  in  the  equation  x  = 
123456789.  Ans.  x  =  8-6400268. 

Ex.   13.    Given  2a:  -  "Ix^  '\-  \\x^  ^  ^x  —  \\ ,  to  find  x. 
Ex.   14.    To  find  the  value  of  x  in  the  equation 

(3x2  —  2  ^  a:  -f  1)«  —  (x2  —  4x  ^  X  -f  3  ^x^^  =  66. 

Ans.  X  =  18-360877. 


To  resolve  Cubic  Equations  hy  Cardenas  Rule, 
I 
Though  the  foregoing  general  method,  by  the  application 
of  Double  Position,  be  the  readiest  way,  in  real  practice,  of 
finding  the  roots  in  numbers  of  cubic  equations,  as  well  as  of 
all  the  higher  equations  universally,  we  may  here  add  the 
particular  method  commonly  called  Garden's  Rule,  for  re- 
solving cubic  equations,  in  case  any  person  should  choose 
occasionally  to  employ  that  method. 

The  form  that  a  cubic  equation  must  necessarily  have  to 
be  resolved  by  this  rule,  is  this,  viz.  z^  -{-  az  =  b,  that  is, 
wanting  the  second  term,  or  the  term  of  the  2d  power  z^. 
Therefore,  after  any  cubic  equation  has  been  reduced  down 
to  its  final  usual  form,  x^  +  px^  -f-  qx  =  r^  freed  from  the 
co-eflicient  of  its  first  term,  it  will  then  be  necessary  to  take 
away  the  2d  term  px^  ;  which  is  to  be  ^one  in  this  manner  : 
Take  ^p,  or  ^  of  the  coefficient  of  the  second  term,  and 
annex  it,  with  the  contrary  sign,  to  another  unknown  letter 
Zy  thus  z  —  i  p  ;  then  substitute  this  for  x,  the  unknown  letter 
in  the  original  equation  x^  -|-  px'^  -{-  qx  =  r^  and  there  will 
result  this  reduced  equation  z^  -\-  az  =  b,  of  the  form  projser 
for  applying  the  following,  or  Garden's  rule.  C^r  take 
c  =  ia,  and  d  =  ^b,  by  which  the  reduced  equation  takes 
this  form  2  3  -I-  3c  /=  2d, 

Thejr^ 


CUBIC,  &c.  EQUATIONS.  26JSI 

Then  substitute  the  values  of  c  and  d  in  this 

form  z  =  ^ci  +  y  (d^  -j- c^)  -f  ^d^^(^d^  +  c^). 


or.  =  3/d -f  v^(ci^  ^c3)-777^-^^^ 

and  the  value  of  the  root  .,  of  the  reduced  equation  2^  + 
az  =  b^  will  be  obtained.  Lastly,  take  x  =  z  —  ip,  which 
will  give  the  value  of  a;,  the  required  root  of  the  original  equa- 
tion x^  4*  px^  +  9^  =  »'i  first  proposed. 

One  root  of  this  equation  being  thus  obtained,  then  de- 
pressing the  original  equation  one  degree  lower,  after  the  man- 
ner described  p  260  and  261,  the  other  two  roots  of  that  equa- 
tion will  be  obtained  by  means  of  the  resulting  quadratic 
equation. 

JVote.  When  the  co-efficient  a,  or  c,  is  negative,  and  c^  is 
greater  thanc?^^  this  is  called  the  irreducible  case,  because 
then  the  solution  cannot  be  generally  obtained  by  this  rule. 

Ex.  To  find  the  roots  of  the  equation  x^—6x^  +  10a;  =  8. 

First,  to  take  away  the  2d  term,  its  co- efficient  being—  6, 
its  3d  part  is  —  2  ;  put  therefore  x  =  .  +  2  ;  then 
x^  =  z^  -{-  6.2  ^  i2z  -{-8 
—  6x2  =       —  6.2  —  24.  ■—  24 
+  lOx  =  4-  10.  4-  20 


theref.  the  sum.  .»      i^        —   2.  +  4  =  8 
or  .3      ♦        —   2.  =  4 
Here  then  a  =  —  2,  6  =  4,  c  =  —  |,  rf  =  2. 


Theref.  l/d^^{d^  4.c3)=3^24-v^(4-^\)=V2-t-v^Vo/-= 

V2  4-  V"  x/  3  =  1-57785 

and  3/d-y(^+c3)=  3/2— ^(4-^\)=^24VVV'= 
3/2— V"  ^/^=  0-42265 
then  the  sum  of  these  two  is  the  value  of  .  =  2. 
Hence  a:  =  .  4"  2  e=:  4,  one  root  of  a;  in  the  eq.  x^  —  6a;2  -^ 
lOx  =  8. 
To  find  the  two  other  roots,  perform  the  division,  &c.  as  in 
p.  261,  thus  : 

a:  —  4)  x3  —  6a;3  -|-  lOa:  —  8  {x^  —2a;  +  2  =  0. 

^3    4a;2 


—  2x2  +  lOx 

—  2x2  +     8x 

2x- 

2x  — 

-8 
-8 

Vot.  I.  3i  Hence 


266  ALGEBRA. 

Hence  x^  —2a?  =—2,  or  a;2  —  2a:  -f  I  =  ~  1,  and  a:  —  I 
=  ^:  v^  -  1  ;  ^  =  1  +  v^  ~-  1  or  =  1  —  v^  —  1,  the  two 
ether  sought. 

Ex.  2.  To  find  the  roots  of  x^  —  9x2  +  28x  =  30. 

Ans.  a:  =  3,  or  =  3  H-  ^  —  1,  or  =  3  —  ^  —  1. 
Ex.  3.  To  find  the  roots  of  x^  -  Tx^  +  Hot  =  20, 

Ans.  a;  =  6,  or  =  1  +  ^  —  3,  or  =  1   -  ^  —  3. 


OF  SIMPLE  INTEREST. 

As  the  interest  of  any  sum,  for  any  time,  is  directly  pro- 
portional to  the  principal  sum,  and  to  the  time  ;  therefore  the 
interest  of  1  pound,  for  1  year  being  multiplied  by  any  given 
principal  sum,  and  by  the  time  of  its  forbearance,  in  years  and 
parts,  will  give  its  interest  for  that  time.  That  is,  if  there 
be  put 

r  =  the  rate  of  interest  of  1  pound  per  annum, 
p  =  any  principal  sum  lent, 
t  =  the  time  it  is  lent  for,  and 

d  =  the  amount  or  sum  of  principal  and  interest  ;  then 
is  prt  =  the  interest  of  the  sum  /?,  for  the  time  t,  and  conseq, 
p  -\-  prt  oTp  X  (1  -{-  rt)  =  a,  the  amount  for  that  time,    i 

From  this  expression,  other  theorems  can  easily  be  deduced, 
for  finding  any  of  the  quantities  above  mentioned  ;  whiph  theo- 
rems; collected  together,  will  be  as  below  : 
1st,  a  =  /)  -{-  prty  the  amount. 
a 

2d,  p  =  ,  the  principal, 

1  +  ri 
a—p 

3d,  r  = ,  the  rate, 

pt 
a—p 

4th,  <  = the  time. 

pr 
For  Example.     Let  it  be  required  to   find,  in  what  time 
any  principal  sum  will  double  itself,  at  any  rate  of  simple  in- 
terest. 

In  this  case,  we  must  use  the  first  theorem,  a  =  p  -{-  prt, 
in  which  the  amount  a  must  be  made  =  2/?,  or  double  the 
principal,  that  is,  p  -|-  prt  =  2p,  or  prt  =  p,  or  rt  =  1  ; 

1 
and  hence  t  =  -. 

r  Here, 


COMPOUND  INTEREST.  267 

Here,  r  being  the  interest  of  IL  for  1  ^^ear,  it  follows,  that 
the  doubling  at  simple  interest,  is  equal  to  the  quotient  of 
any  sum  dirided  by  its  interest  for  1  year.  So,  if  the  rate  of 
interest  be  6  per  cent,  than  100  -i-  5  =  20,  is  the  time  of 
doubling  at  that  rate. 

Or  the  4th  theorem  gives  at  once 
a-p       2p— /)       2— 1         1 

t  = = = s=  — ,  the  same  as  before. 

pr  pr  r  r 


COMPOUND  INTEREST. 

Besides    the    quantities    concerned    in    Simple     Interest 
namely, 

p  =  the  principal  sum, 

r  =  the  rate  or  interest  of  11.  for  1  year, 

a  =  the  whole  amount  of  the  principal  and  interest, 

t  =  the  time, 
there  is  another  quantity  employed  in  Compound  Interest, 
viz.  the   ratio  of  the  rate  of   interest,  which  is  the  amount 
of  1/.  for  1  time  of  payment,  and  which  here  let  be  denoted 
by  R,  viz. 

R  =  1  -}-  r,  the  amount  of  1/.  for  1  time. 
Then  the  particular  amounts  for  the  several  times  may  be 
thus  computed,  viz.     As  1/.  is  to  its  amount  for  any  time  su  is 
any  proposed  principal  sum,  to  its  amount  for  the  same  time  ; 
that  is,  as 

11.   :  R  :  :  p        :  pR,    the  1st  year's  amount, 

11.   :   R  :  :  pR     :  pR^ ,  the  2d  year's  amount, 

1/.  :  R  :  :  pR^   ;  pR^,  the  3d  year's  amount, 

and  so  on. 
Therefore,  in  general,  pR*  =  a  is  the  amount  for  the  i  year, 
or  t  time  of  payment.     Whence  the  following  general  theoi» 
reras  are  deduced  : 

1st,  a  =  joRt,  the  amount, 
a 

2d,  /?  =  —  the  principal, 

a 
3d,  R  =  ^  — ,  the  ratio, 

P 
log.  of  a  —  log.  of  p 

4th,  t  = ,  the  time. 

lo£.  of  R 

From 


^6d 


ALGEBRA. 


From  which,  any  one  of  the  quantities  may  be  found,  wheji 
the  rest  are  given. 

As  to  the  whole  interest,  it  is  found  by  barely  subtracting 
the  principal  p  from  the  amount  a. 

Example.  Suppose  it  be  required  to  find,  in  how  many 
years  any  principal  sum  will  double  itself,  at  any  proposed 
rate  of  compound  interest. 

In  this  case  the  4th  theorem  must  be  employed,  making 
(0  =  2p  ♦  and  then  it  is, 

'     log.  a — log./)       log.  2p  — log. /)       log.  2 

log.  R  log.  R  log.  R 

So,  if  the  rate  of  interest  be  5  per  cent,  per  annum  ;  then 
^R  =  1  -f  -06  =  1  -05  ;  and  hence 
log.  2         -301030 

t  —  -' —  = =  14-2067  nearly  ; 

log.  106      -021189 
that  is,  any  sum  doubles  itself  in  14a  years  nearly,  at  the  rate 
of  6  per  cent,  per  annum  compound  interest. 

Hence,  and  from  the  like  question  in  Simple  Interest, 
above  given,  are  deduced  the  times  in  which  any  sum  doubles 
itself,  at  several  rates  of  interest,  both  simple  and  com- 
pound ;  viz. 


At-, 


per  cent,  per  annum 
interest,    1/.  or  any 

other  sum,  will 
double  itself  in  the 

following  years. 


"  At  Simp.  Int. 

At  Comp.  Int. 

in  50 

in  35-0028 

40 

28-0701 

33i 

23-4498 

284 

20-1488 

25      «1 

1 7-6730  «5 

22f    S 

15-7473  S 

20      ? 

14-2067?' 

16| 

11-8967 

14f 

10-2448 

12i 

9-0065 

IH 

8-0432 

L         10 

1          7  2725 

The 


eOMPOTJND  INTEREST. 


f69 


Th€  following  Table  will  very  much  facilitate  calculations 
•f  compouad  interest  on  any  sum,  for  any  number  ot  years, 
at  various  rates  of  interest. 


The  Amounts  of 

\l.  in  an) 

r  Number  of  Year« 

. 

YTi: 

3 

H 

4 

H 

5 

6 

1 

1-0:^00 

1-0350 

1  ()4i-0 

1-04.^0 

1  0500 

1  0600 

2 

1-0609 

J -0712 

10816 

1-0920 

1102o 

1  1^36 

3 

10927 

1-1087 

11249 

1.1412 

1-1576 

M910 

4 

11255 

11475 

1-1699 

11925 

i-2155 

1-2625 

5 

1  1593 

1-1877 

1-2167 

1-2462 

1-2763 

1-3382 

6 

11941 

1-2293 

1-2653 

1  -302^ 

1-3401 

1-4185 

7 

1  2:^99 

1-2723 

1-3159 

1  36U9 

1-4071 

1-5036 

8 

1-2668 

1  3168 

1-3686 

,  4221 

1-4775 

1-5939 

9 

1  3048 

1-3629 

1-4233 

1-4861 

1-5513 

1  -6895 

10 

1-3439 

1-4106 

1-4802 

1-5530 

1-628^ 

1-7909 

11 

1-3842 

1-4600 

1-5895 

1-6229 

1-710;': 

1-8983 

12 

14258 

1-6111 

1  6010 

1-6959 

1-7950 

2-0122 

13 

1-4685 

1-5640 

1-66^1 

1  7722 

1  -8866 

2  1329 

14 

1-6126 

1-6187 

1-7317 

1-8519 

19799 

2-2609 

15 

1-5580 

1-6753 

1-8009 

1-9353 

2-0789 

2-3966 

16 

1-6047 

1-7340 

1-8730 

2-0224 

2-1829 

2-6404 

17 

1-6528 

1-7947 

1-9479 

2-1134 

2-2920 

2-6928 

18 

1-7024 

1-8575 

2-0258 

2-2085 

2-4066 

2-8543 

19 

1-7535 

1-9225 

2-1068 

2-3079 

2-5270 

3 -02^6 

20 

1  1-8061 

1-9898 

2-1911 

2-4117 

2-6533 

,3-2071 

The  use  of  this  Table,  which  contains  all  the  powers,  r*, 
to  the  20th  power,  or  the  amounts  of  1/.  is  chiefly  to  cal- 
culate the  interest,  or  the  amount  of  any  principal  sum,  for 
any  time,  not  more  than  20  years. 

For  example,  let  it  be  required  to  find,  to  how  much  523/. 
will  amount  in  15  years,  at  the  rate  of  5  per  cent,  per  annum 
/compound  interest. 

In  the  table,  on  the  line  15,  and  in  the  column  5  per  cent, 
is    the    amount  of  1/.    viz.         -         -         2-0789 
this  multiplied  by  the  priDcipai         -  523 

gives  the  amount         -         -         -  1087-2647 

or  -         -         -  1087/.  5s.  S^d, 

and  therefore  the  interest  is  -  664/.  5s.  3\d. 

Note  1.    When  the    rate  of  interest  is  to  be  determined  to 

any  other  time  than  a  year  ;  as   suppose  to  i  a  year,  or  ^  a 

year,   &c.  \   the  rules  are  still  the  same  ;    but  then  t  will 

express 


27^.  ALGEBRA.       = 

express  that  time,  and  r  must  be  taken  the  amount  for  that 
time  also. 

JsToie  2.  When  the  compound  interest,  or  amount,  of  any 
sura  is  required  for  the  parts  of  a  jear  ;  it  may  be  deter- 
mined in  the  following  manner  : 

\st.  For  any  time  which  is  some  aliquot  part  of  a  year  : — 
Find  the  amount  of  1/.  for  1  year,  as  before  ;  then  that  root 
of  it  which  is  denoted  by  the  aliquot  part,  will  be  the  amount 
of  IL  This  amount  being  multiplied  by  the  principal  sum, 
will  produce  the  amount  of  the  given  sum  as  required. 

2{/,  When  the  time  is  not  an  aliquot  part  of  a  year  : — 
Reduce  the  time  into  days,  and  take  the  366tii  root  of  the 
amount  of  \l.  for  1  year,  which  will  give  the  amount  of  the 
same  for  1  day.  Then  raise  this  amount  to  that  power  whose 
index  is  equal  to  the  number  of  days,  and  it  will  he  the  amount 
for  that  time.  Which  amount  being  multiplied  by  the  princi- 
pal sum,  will  produce  the  amount  of  that  sum  as  before.—  And 
in  these  calculations,  th^  operation  by  logarithms  will  be 
Tery  useful. 


OF  ANNUITIES. 


ANNUITY  is  a  term  used  for  any  periodical  income,  aris- 
ing from  money  lent,  or  from  houses,  lands,  salaries,  pen- 
sions. &c.  payable  from  time  to  time,  but  mostly  l?y  annual 
payments. 

Annuities  are  divided  into  those  that  are  in  Possession,  and 
those  in  Reversion  :  the  former  meaning  such  as  have  com- 
menced ;  and  the  latter  such  as  will  not  begin  till  some  par- 
ticular event  has  happened,  or  till  after  some  certain  time  has 
elapsed. 

When  an  annuity  is  forborn  for  some  years,  or  the  pay- 
ments not  made  for  that  time,  the  annuity  is  said  to  be  in 
Arrears. 

An  annuity  may  also  be  for  a  certain  number  of  years  ; 
or  it  may  be  without  any  limit,  and  then  it  is  called  a  Per- 
petuity. 

The  Amount  of  an  annuity,  forborn  for  any  number  of 
years,  is  the  sum  arising  from  the  addition  of  all  the  annui- 
ties for  that  number  of  years,  together  with  the  interest  due 
upon  each  after  it  bqcwnes  d.ue. 

The 


ANNUITIES.  i'71 

The  Present  Worth  nr  Vahie  of  an  annuity,  is  the  price  or 
sum  wliich  ought  to  .e  given  for  it,  supposing  it  to  be  bought 
off,  or  paid  all  at  once. 

Let  a  =  the  annuity,  pension,  or  yearly  rent ; 
n  =  the  number  of  years  forborn,  or  lent  for  ; 
R  =  the  aiTiount  of  ll.  for  1  year  ; 
fn  =  the  amount  of  the  annuity  ; 
V  =  its  value,  or  its  present  worth. 
Now,  1  being  the  present  value  of  the  sum  r,  by  propor- 
tion the  present  value  of  any  other  sum  a,  is  thus  found  : 

a 
as  R  :  1  :  :  a  :  —  the  present  value  of  a  due  1  year  Iienc3. 


In  hke  mannner  —  is  the   present  value  of  a  due  2  years 

R2 

ct      a  a      a     a 

hence  ;  for  r  :  1    :  :  —  :  —     So  also  — ,  — ,  — ,   &c.    will 

R         R2  r3        r.4      r5 

be  the  present  values  of  a,  due  at  the  end  of  3,  4,  6,  &c. 
years  respectively.  Consequently  the  sum  of  all  these,  or 
a         a        a         a  1111 

—  +  --  +  —  +  — 4-&C.  =  (—  +  —  +  — +-&c.)X  a, 

R  r2  r3  r4  R  R2  r3  ^4 

continued  to  n  terms,  will  be  the  present  value  of  all  the  n 
years'  annuities.  And  the  value  of  the  perpetuity,  is  the  sum 
of  the  series  to  infinity. 

But  this  series,  it  is  evident,  is  a  geometrical  progressi(Mi, 
1 
having  —  both  for  its  first  term  and  common  ratio,  and  the 

ft 
number  of  its  terms  n  ;  therefore  the  sum  u  of  all  the  terms 
©r  the  present  value  of  all  the  annual  payments,  will  be 
11        1 

R        R       R"                        r"  —  1         a 
•»  = X  a,  or  = X  — . 

1  R   —    1  R" 

1  — 

R 

When  the  annuity  is  a  perpetuity  ;  n  being  infinite,  r"  ib 

1 
also  idfinite,  and  therefore  the  quantity  —  becomes   =   0. 

R" 

a  1 

therefore X  —  also  =  0  ;  consequently  the  expression 

R-  I         R» 

becOHie^ 


27g  ALGEBRA. 

a 
becomes  barely  v  = ;  that  is,    any  annuity  divided  by 

R-l 

tbe  interest  of  \l  for  1  year,  gives  the  value  of  the  perpetuity. 
So,  if  the  rate  of  interest  be  6  per  cent, 

Then  lOOa  -r-  5  =  20rt  is  the  value  of  the  perpetuity  at 
£  per  cent  :  Also  lOOa  -f-  4  =  25a  is  the  value  of  the  per- 
petuity at  4  per  cent  :  And  100a  -J-  3  =  33^a  is  the  value  of 
the  perpetuity  at  3  per  cent :  and  so  oa. 

Again,  because  the  anriount  of   1/.    in  n  years,  is  r",  its 

increase  in  that  time  will  be  r'*  —  1  ;  but  its  interest  for  one 

iingle  year,   or  the  annuity  answering   to  that  increase,    is 

R  —  1  ;  therefore  as  r  —  1  is  to  r"  —  1,  so  is  a  to  m  ;  that 

R"  —  1 

is,  m  =  X  a.     Hence,  the  several  cases  relating  ta 

R  —    1 
Annuities  in  Arrear,  will  be  resolved  by  the  following  equa- 
feons  : 

R»-~l 

■m  = '—  X  a  =  -VR^  I 


R  

1 

R"- 

1 

X— : 

R» 

m 
5 

R« 

R    

1 

R   

1 

Xm 

R 



1 

X  VK*; 

R«-- 

1 

R° 



1 

mR  —  w 

4- 

tb 

log.  m  —  log.  "& 

n  = ■ 

log.  R 
log.  m  —  log.  t 


Log.  R  = 


1       1 


RP         R"         R    — -    1 

In  this  last  theorem,  r  denotes  the  present  value. of  aa 
annuity  in  reversion,  after/)  years,  or  not  commencing  till 
after  the  first  p  years,  being  found  by  taking  the  difference 
r"  — ■  1        a  rP  —  1        a 

between  the  two  values X  —  and —7  X  — ,       for 

R  —  1       r"         r   —  1       rP 
n  years  and  p  years. 

But  the  amount  and  present  vaUie  of  any  annuity  for  any 
number  of  years,  up  to  21,  will  be  most  readily  found  by  the 
tno  following  tables.  tablk 


ANNUITIES. 


27a 


TABLE  I. 
The  Amount  of  an  Annuity  of  ll.  at  Compound  Interest 


YrvjatSperc. 

3)  pe:  < ..  4  !>er  r 

4iperr.| 

5  rei  r. 

6  p'  re 

1 

10000 

10000 

10000 

1-0000 

1  (000 

1  OCOC 

2 

2  0300 

2  0350 

20400 

2  04.0 

2  0500 

2  Or.  00 

3 

3  0909 

3  3062 

3  1216 

3- 1370 

3  35:;5 

3  1836 

4 

4  18.36 

4  2l4y 

4  :>465 

4  4782 

43101 

4  3746 

5 

5  3091 

5-3625 

5-4»6.. 

54707 

5-5-ot 

56371 

6 

6-4684 

6  5502 

6  6330 

6  7169 

6  8019 

6  9.53 

7 

7-66 -'5 

7-77i'4 

7  8983 

8-0192 

8  1420 

8  3938 

8 

8-8923 

9  0517 

9  142 

9  3800 

9-5191 

9  8975 

9 

10  1591 

10  3685 

105828 

108021 

11-0266 

li  4913 

10 

11  4639 

n  73i4 

12  006! 

12-2882 

12-5779 

13  1808 

U 

12  8078 

13  1420 

13  48  4 

13  8412 

14  2068 

149716 

12 

14  1920 

14  60^<0 

15  0258 

15  4640 

159171 

16  8699 

13 

156178 

\6  1130 

^  6  6268 

17  :5y9 

17  7130 

18  8821 

14 

17*0863 

17  6770 

18  2919 

189321 

19  5986 

21  0.51 

15 

18  5989 

19  2957 

20  3236 

';0  7841 

21 5786 

23  2760 

16 

20-1569 

20  9710 

I  8245 

2.7193 

23  6o75 

25  67^5 

17 

217616 

227050 

23  6975 

^4  7417 

25-8404 

28-2129 

18 

234141 

24  4997 

25-6454 

26-8551 

28- 1324 

30  9057 

19 

25  1169 

26  357 

27-6712 

290636 

SO  5390 

337600 

20 

268704 

58-2797 

29-7781 

31  3714 

33  0660 

36  7856 

21 

28-6/65 

30  2695 

31  9692 

33-7831 

S5  7193 

39  99  ?7 

TABLE  II.    The  present  value  of  an  Annuity  of  1^ 


^K 

at  3perc 

33Pvrc. 

4!ier  c. 

4^  per  c 

5  jjcrc. 

6  perc. 

1 

0  9709 

0-9662 

0  9615 

0  9569 

0  9524 

0-9434 

2 

1  9135 

18997 

18861 

1-8727 

1  8594 

1-8334 

3 

2  8286 

28016 

2  7751 

2-7490 

27233 

2-6730 

4 

3-7171 

36731 

3  6299 

3  5875 

35460 

3  4651 

5 

4  5797 

45151 

4-4518 

4  5900 

4-3295 

4-2124 

6 

5  4172 

53286 

5  2421 

51579 

50757 

49173 

7 

62303 

6  1*45 

600-0 

5-8927 

5  7864 

55824 

8 

7  0197 

6-8740 

67327 

6  5959 

6  4632 

6-^^098 

9 

7  7861 

7  6077 

74353 

7-2688 

7  1078 

68017 

10 

8  5302 

8-3166 

8  1109 

7  9127 

7-7217 

7  3601 

U 

9  2526 

,  9  0116 

8  7605 

8  5289 

8-3054 

7  8869 

12 

9-9540 

9  6633 

9  3851 

9  1186 

88633 

83838 

13 

10  6^>50 

10-3027 

9  9857 

9  6829 

9  3936 

885/7 

14 

11  296  i 

10  9205 

10  5631 

10  2228 

9  8986 

9  2950 

15 

1 1-9379 

,115174 

111184 

10  7396 

'0  3797 

9  712^ 

16 

12  5611 

12  0941 

116523 

11  2340 

10  8378 

10  10  59 

17 

13  1661 

126513 

12  1657 

117072 

11  2741 

10  4773 

18 

13  7535 

13  1897 

126593 

12-1600 

11-6896 

10  8276 

19 

14  3238 

13  7098 

13  1339 

12*5933 

12  0853 

11 1581 

20 

14  8775 

l4-oi24 

135903 

13  0079 

12-4622 

114699 

21 

154150 

14  6980 

140292 

13  4047 

'  12-8212 

11  7641 

Vol.  L 


36 


n 


274  ALGEBRA. 

To  find  the  Amount  of^  any  annuity  forborn  a  certain  number 
of  years. 

Take  out  the  amount  of  \L  from  the  first  table,  for  the 
proposed  rate  and  time  ;  then  multiply  it  hy  the  given  annuity; 
and  the  product  will  be  the  amount,  for  the  same  number  of 
years,  and  rate  of  interest. — And  the  converse  to  find  the  rate 
or  time.  y 

Exam.  To  find  how  much  an  annuity  of  60Z.  will  amount  to 
in  20  years,  at  3^  per  cent,  compound  interest. 

On  the  line  of  20  years,  and  in  the  column  of  3i  per  cent, 
stands  28*2797,  which  is  the  amount  of  an  annoity  of  \l.  for 
ihe  20  years.  Then  28-2797  X  50  gives  1413'986/.  = 
iW13/.  19s.  8d.  for  the  answer  required. 

To  find  the  present  Value  of  any  annuity  for  any  number  of 
years. — Proceed  here  by  the  2d  table,  in  the  same  manner  as 
above  for  the  1st  table,  and  the  present  worth  required  will 
be  found. 

Exam.  1.  To  find  the  present  value  of  an  annuity  of  501, 
which  is  to  continue  20  years,  at  3^  per  cent. — By  the  table, 
the  present  value  of  IZ,  for  the  given  rate  and  time,  is 
14-2124  ;  therefore  14-2124  X  50  =  710-6,2/.  (yr  710Z.  126-.  4d, 
is  the  present  value  required. 

Exam,  2.  To  find  the  present  value  of  an  annuity  of  20/. 
to  commence  10  years  hence,  and  then  to  continue  for  1 1 
years  longer,  or  to  terminate  21  years  hence,  at  4  per  cent, 
interest. — in  such  cases  as  this,  we  have  to  find  the  difierence 
between  the  present  values  of  two  equal  annuities,  for  the 
two  given  times  ;  which  therefore  will  be  done  by  subtract- 
ing the  tabular  value  of  the  one  period  from  that  of  the 
other,  and  then  multiplying  by  the  given  annuity.  Thus, 
tabular  value  for  21  years  14*0292 
ditto  for      -     -     10  years    8-1109 


the  difference  5-9183 
multiplied  by  20 


gives       -        118-366/. 

or       -        -  118/.  7s.  3^d.  the  answer. 


END  (IF  THE  ALGEBRA. 


GEOMETRY. 


DEFINITIONS. 


1.  -A.  POINT  is  that  wh^ch  has  positioa, 
but  no  magnitude,  nor  dimensions  ;  neither 
length,  breadth,  nor  thickness. 

2.  A  Line  is  length,  without  breadth  or 
thickness. 

3.  A  Surface  or  Superficies,  is  an  extension 
or  a  figure,  of  two  dimensions,  length  and 
breadth  ;  but  without  thickness. 

4.  A  Body  or  Solid,  is  a  figure  of  three  di- 
mensions, namely,  leagth,  breadth,  and  depth, 
or  thickness. 

6.  Lines  are  either  Right,  or  Curved,  or 
Mixed  of  these  two. 

6.  A  Right  Line,  or  Straight  Line,  lies  all 
in  the  same  direction,  between  its  extremities  ; 
and  is  the  shortest  distance  between  two  points. 

When  a  line  is  mentioned  simply,  it  means 
a  Right  Line. 

7.  A  Curve  continually  changes  its  direction 
between  its  extreme  points. 

8.  Lines  are  either  Parallel,  Oblique,  Per- 
pendicular, or  Tangential. 

9.  Parallel  Lines  are  always  at  the  same  per- 
pendicular distance  ;  and  they  never  meet  though 
ever  so  far  produced. 

10.  ObUque  lines  change  their  distance,  and 
would  meet,  if  produced  on  the  side  of  the 
least  distance. 

11.  One  line  is  Perpendicular  to  another, 
when  it  inclines  not  more  on   the    one  side 

than 


(f7 

Jrd7 


L/ 


276  GEOMETRY, 

than  the   other,  or  when  the    angles  on  both 
sides  of  it  are  equal. 

12.  A  line  or  circle  is  Tangential,  or  a 
Tangent  to  a  circle,  or  other  curve,  when  it 
touches  it,  without  cutting,  when  both  are  pro- 
duced. 

13.  An  Angle  is  the  inclination  or  opening 
of  two  lines,  having  different  directions,  and 
meeting  in  a  point. 

14.  Angles  are  Rigkt  or  Oblique,  Acute  or 
©btuse. 

15.  A  Right  Angle  is  that  which  is  made  by 
one  line  perpendicular  to  another.  Or  wheii 
the  angles  on  each  side  are  equal  to  one  an- 
other, they  are  right  angles. 

16.  An  Oblique  Angle  is  that  which  is 
made  by  two  oblique  lines  ;  and  is  either  less 
or  greater  than  a  right  angle. 

17.  An  Acute  Angle  is  less  than  a  right 
angle. 

18.  An  Obtuse  Angle  is  greater  than  a  right 
angle. 

19.  Superficies  are  either  Plane  or  Curved. 

20.  A  Plane  Superficies,  or  a  Plane,  is  that  with  which 
a  right  line  may,  every  way  coincide.  Or,  if  the  line  touch 
the  plane  in  two  points,  it  will  touch  it  in  every  point.  But, 
if  not,  it  is  curved. 

21.  Plane  figures  are  bounded  either  by  right  lines  or 
curves. 

22.  Plane  figures  that  are  bounded  by  right  lines  have 
names  according  to  the  number  of  their  sides,  or  of  their 
angles  ;  for  they  have  as  many  sides  as  angles  ;  the  least 
number  being  three. 

23.  A  figure  of  three  sides  and  angles  is  called  a  Triangle. 
And  it  receives  particular  denominations  from  the  relations 
of  its  sides  and  angles. 

24.  An  Equilateral  Triangle  is  that  whose 
three  sides  are  all  equal. 


A 

26.  An  Isosceles  Triangle  is  that  which  has  /  \ 

two  sides  equal.  /y^ 


DEFINITIONS. 


277 


26.  A  Scalene  Triangle  is  that  whose  three 
sides  are  all  enequal. 

27.  A    Right-angled    Triangle  is  that  which 
has  one  ritcht-angle 

28.  Other  triangles  are  Oblique-angled,  and 
are  either  Obtuse  or  Acute 

29.  An  Obtuse-angled   Triangle  has  one  ob- 
tuse angle. 

30.  An  Acute-angled  Triangle  has  all  its  three 
angles  acute. 


31.  A  figure  of  Four  sides  and  angles  is  called 
a  Quadrangle,  or  a  Quadrilateral. 

32.  A  Parallelogram  is  a  quadrilateral  which 
has  both  its  pairs  of  opposite  sides  parallel. 
And  it  takes  the  following  particular  names, 
viz.  Rectangle,  Square,  Rhombus,  Rhomboid. 

33.  A  Rectangle  is  a  parallelogram  having  a 
right  angle. 

34.  A  Square  is  an  equilateral  rectangle  ; 
having  its  length  and  breadth  equal. 

35.  A  Rhomboid  is  an  oblique-angled  paral- 
lelogram. 


36.  A  Rhombus  is  an  equilateral  rhomboid  ; 
having  all  its  sides  equal,  but  its  angles  ob- 
lique. 

37.  A  Trapezium  is  a  quadrilateral  which 
hath  not  its  opposite  sides  parallel. 

38.  A  Trapezoid  has  only  one  pair  of  oppo- 
site sides  parallel.  i         V 

39.  A  Diagonal  is  a  line  joining  any  two  op- 

posite  angles  of  a  quadrilateral.  /N.      I 

40^  Plane  figures  that  have  more  than  four  sides,  are,  in 

general,  called   Polygons  :  and  they  receive  other  particular 

names,   according  to  the  number    of  their  sides  or  angles 

^  Thus,  * 

41.  A  Pentagon  is  a  polygon  of  five  sides  :  a  Hexagon,  of 
six  sides  ;  a  Heptagon,  seven ;  an  Octagon,  eight ;  a  Nona- 
gon,  nine;  a  Decagon,  ten  ;  an  Undecagoa,  eleven  ;  and  a  Do- 
decagon, twelve  sides. 


42.  A 


27S  GEOMETRY. 

42.  A  Regular  Polygon  has  all  its  sides  and  all  its  angles 
equal. — If  they  are  not  both  equal,  the  polygon  is  Irregular. 

43.  An  Equilateral  Triangle  is  also  a  Regular  Figure  of 
three  sides,  and  the  Square  is  one  of  four  ;  the  fornjer  being 
also  called  a  i  rigon,  and  the  latter  a  Tetragon. 

44.  Any  figure  is  equilateral,  when  all  its  sides  are  equal  : 
and  it  is  equiangular  T\hen  all  its  angles  are  equal.  When 
both  these  are  equal,  it  is  a  regular  figure. 

46.  A  Circle  is  a  plain  figure  bounded  by 
a  curve  line,  called  the  Circnuiference,  which 
is  every  where  equidistant  from  a  certain  point 
within,  called  its  Centre. 

The  circumference  itself  is  often  called  a 
circle,  and  also  the  Periphery. 

46.  The  Kadius  of  a  circle  is  a  line  drawn 
from  the  centre  to  the  circumference. 


47.  The  Diameter  of  a  circle  is  a  line  drawn 
through  the  centre,  and  terminating  at  the  cir- 
cumference on  both  sides. 


48.  An  Arc  of  a  circle  is  any  part  of  the 
circumference. 


49.    A  Chord  is  a  right  line  joining  the  ex- 
tremities of  an  arc. 


50.  A  Segment  is  any  part  of  a  circle  bound- 
ed by  an  arc  and  its  chord. 

61.  A  Semicircle  is  half  the  circle,  or  a  seg- 
ment cut  off  by  a  diameter. 

The  half  circumference  is  sometimes  called 
the  semicircle. 

62.  A  Sector  is  any  part  of  a  circle  which 
is  bounded  by  an  arc,  and  two^adii  drawn  to  its 
extremities. 

63.  A  Quadrant,  or  Quarter  gf  a  circle,  is  a 
sector  having  a  quarter  of  the  circumference 
for  its  arc,  and  its  two  radii  are  perpendicular 
to  each  other.  A  quarter  of  the  circumference 
is  sometimes  called  a  Quadrant. 


64.  Tl^,e 


DEFINITIONS. 


27S 


54.  The  Height  or  Altitude  of  a  6gure  is 
a  perpendicular  let  fall  from  ah  angle,  or  its 
vertex,  to  the  opposite  side,  called  the  base. 

55.  In  a  right-augled  triangle,  the  side  op- 
posite the  right  angle  is  called  the  Hypothe- 
nuse  ;  and  the  other  two  sides  are  called  the 
Legs,  and  sometimes  the  Base  and  Perpendi- 
cular. 

56.  When  an  angle  is  denoted  by  three 
letters,  of  which  one  stands  at  the  angular 
point,  and  the  other  two  on  the  two  sides, 
that  which  stands  at  the  angular  pomt  is  read 
in  the  middle.  Thus  the  angle  contained 
by  the  lines  BA  and  AD  is  called  the  angle 
BAD  or  DAB. 

57.  The  circumference  of  every  circle  is 
supposed  to  be  divided  into  360  equal  parts, 
called  Degrees  :  and  each  degree  into  60  Mi- 
nutes, each  minute  into  60  Seconds,  and  so  on. 
Hence  a  seniicircle  contains  1 80  degrees,  and 
a  quadrant  90  degrees. 

58.  The  Measure  of  an  angle,  is  an  arc  of 
any  circle  contained  between  the  two  lines 
which  form  that  angle,  the  angular  point  being 
the  centre  ;  and  it  is  estimated  by  the  num- 
ber of  degrees  contained  in  that  arc. 

59.  Lines,  or  chords,  are  said  to  be  Equi- 
distant from  the  centre  of  a  circle,  when  per- 
pendiculars drawn  to  them  from  the  centre 
are  equal. 

60.  And  the  right  line  on  which  the  Great- 
er Perpendicular  falls,  is  said  to  be  farther 
from  the  centre. 

61.  An  Angle  In  a  segment  is  that  which 
is  contained  by  two  lines,  drawn  from  any 
point  in  \he  arc  of  the  segment,  to  the  two 
extremities  of  that  arc. 

62.  An  Angle  On  a  segment,  or  an  arc,  is  that  which  is 
contained  by  two  lines,  drawn  from  any  point  in  the  opposite 
or  supplemental  part  of  the  circumference,  to  the  extremi- 
ties of  the  arc,  and  containing  the  arc  between  them. 

63.  An  Angle  at  the  circumference,  ifi  that 
whose  angular  point  is  any  wh^re  in  the  cir- 
cumference And  an  angle  at  the  centre,  is 
that  whose  angular  point  is  at  the  centre. 

64.  A 


280 


GEOMETRY. 


64.  A  right-lined  figure  is  Inscribed  in  a 
circle,  or  the  circle  Circumscribes  it,  when 
all  the  angular  points  of  the  figure  are  in  the 
circumference  of  the  circle. 

65.  A  right-lined  figure  Circumscribes  a 
circle,  or  the  circle  is  Inscribed  in  it,  when  all 
the  sides  of  the  figure  touch  the  circumference 
of  the  circle. 

66.  One  right-lined  figure  is  Inscribed  in 
another,  or  the  latter  Circumscribes  the  for- 
mer, when  all  the  angular  points  of  the  for- 
mer are  placed  in  the  sides  of  the  latter. 

67.  A  Secant  is  a  line  that  cuts  a  circle, 
lying  partly  within,  and  partly  without  it. 


68.  Two  triangles,  or  other  right-lined  figures,  are  said  to 
be  mutually  equilateral,  when  all  the  sides  of  the  one  are 
equal  to  the  corresponding  sides  of  the  other,  each  to  each  : 
and  they  are  said  to  be  mutually  equiangular,  when  the  angles 
•f  the  one  are  respectively  equal  to  those  of  the  Other. 

69.  Identical  figures,  are  such  as  are  both  mutually  equ^" 
lateral  and  equiangular  ;  or  that  have  all  the  sides  and  all  the 
angles  of  the  one,  respectively  equal  to  all  the  sides  and  all  the 
angles  of  the  other,  each  to  each  ;  so  that  if  the  one  figure 
were  applied  to,  or  laid  upon  the  other,  all  the  sides  of  the  one 
would  exactly  fall  upon  and  cover  all  the  sides  of  the  other  ; 
the  two  becoming  as  it  were  but  one  and  the  same  figure. 

70.  Similar  figures,  are  those  that  have  all  the  angles  of 
the  one  equal  to  all  the  angles  of  the  other,  each  to  each,  and 
the  sides  about  the  equal  angles  proportional, 

71.  The  Perimeter  of  a  figure,  is  the  sum  of  all  its  sidei 
taken  together. 

72.  A  Proposition,  is  something  which  is  either  proposed 
•to  be  done,  or  to  be  demonstrated,  and  is  either  a  problem  or 

a  theorem. 

73.  A  Problem,  is  something  proposed  to  be  done. 

74.  A  Theorem,  is  something  proposed  to  be  demonstrated. 

75.  A  Lemma,  is  something  which  is  premised,  or  demon- 
strated, in  order  to  render  what  follows  more  easy^ 

76.  A  CoroUory,  is  a  consequent  truth,  gained  immediately 
from  some  preceding  truth  or  demonstration. 

77.  A  Schohum,  is  a  remark  or  observation  made  upcm 
something  going  before  it. 

AXIOMS. 


[  281   1 
AXIOMS. 


1.  Things  which  are   equal  to  the  same  thing  are  equal 
to  each  other. 

2.  When  equals  are  added  to  equals,  the  wholes  are  equal. 

3.  When  equals  are  taken  from  equals,  the  remainders  are 
equal. 

4.  When  equals  are  added  to  unequals,  the  wholes  are 
unequal. 

5.  When  equals  are  taken  from  unequals,  the  remainders 
are  unequal.  ' 

6.  Things  which  are  double  of  the  same  thing,  or  equal 
things,  are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,  are  equal. 

8.  Every  whole  is  equal  to  all  its  parts  taken  together. 

9.  Things  which  coincide,  or  fill  the  same  space,  are  iden* 
tical,  or  mutually  equal  in  all  their  parts. 

10.  All  right  angles  are  equal  to  one  another. 

11.  Angles  that  have  equal  measures,  or  arcs,  are  equal. 


THEOREM  ] 


If  two  Triangles  have  Two  Sides  and  the  Included  Angle 
in  the  one,  equal  to  Two  Sides  and  the  Included  Angle 
in  the  other,  the  Triangles  will  be  Identical,  or  equal  in  all 
respects. 

In  the  two  triangles  abc,  def,  if 
the  side  ac  be  equal  to  the  side  df, 
and  the  side  bc  equal  to  the  side  ef, 
and  the  angle  c  equal  to  the  angle  f  ; 
then  will  the  two  triangles  be  iden- 
tical, or  equal  in  all  respects.  A  B   D        E 

For  conceive  the  triangle  abc  to  be  applied  to,  or  placed 
on,  the  triangle  dbf,  in  such  a  manner  that  the  point  c  may 

Vol.  L  37  coincide 


GEOMETRY. 


coincide  with  the  point  f,  and  the  side  ac  with  the  side  df, 
which  is  equal  to  it. 

Then,  since  the  angle  f  is  equal  to  the  angle  c  (hy  hyp), 
the  side  bc  will  fall  on  the  side  ef.  Also,  because  ac  is 
equal  to  df,  and  bc  equal  to  ef  (by  hyp),  the  point  a  will 
coincide  with  the  point  d,  and  the  point  b  with  the  pt-int  e  ; 
consequently  the  side  ab  will  coincide  with  the  side  de. 
Therefore  the  two  triangles  are  identical,  arid  have  all  th  ir 
other  corresponding  parts  equal  (ax.  9),  namely,  the  ^id;'  ab 
equal  to  the  side  de,  the  angle  a  to  the  angle  d,  and  the 
angle  b  to  the  angle  e.     q.  e.  d, 

THEOREM  n. 

When  Two  Triangles  have  Two  Angleis  and  the  included 
Side  in  the  one.  equal  to  '3'wo  Anjj;les  and  the  included  Side 
in  the  other,  the  Triangles  are  Identical^  or  have  their  olher 
sides  and  angle  equal. 

Let  the  two  triangles    abc,  def,  q  F 

have  the  apgle  a  equal  to  the  angle 
D,  the  angle  b  equal  to  the  angle  e, 
and  the  side  ab  equal  to  the  side  de  ; 
then  these  two  triangles  will  be 
identical. 

For,  conceive  the  triangle  abc  to  be  placed  on  the  trfangle 
DEF.  in  such  nnanner  that  the  side  ab  may  fall  exactly  on  the 
equal  side  de.  Then,  since  the  angle'^A  is  equal  to  the  angle^ 
D  (by  hyp.),  the  side  ac  n«nst  hU  on  the  side  df  ;  and,  in 
like  manner,  because  the  angle  b  is  equal  to  the  angle  e  the 
side  bc  must  fall  on  the  side  ef.  Thus  tbetlree  sides  of  the 
triangle  abc  will  be  exai  tly  placed  on  the  three  sides  of  the 
triangle  def  :  consequently  the  two  triangles  are  identical 
(ax.  9),  having  the  other  two  sides  ac,  bc,  equal  to  ti»e  two 
df,  ef,  and  the  remaining  angle  c  equal  to  the  remaining 
angl^  F.     ci.  E.  D. 

THEOREM  HI. 

In  an  Isos^celes  triangle,  the  Angles  at  the  Base  are  equal. 
Or,  if  a  Triangle  have  Two  Sides  equal,  their  Opposite 
Angles  will  also  be  equal. 

if  the  triangle  abc  have  the  side  ac  equal 
to  the  side  bc  :  then  will  the  angle  b  be 
equal  to  the  angle  a. 

For,  conceive  the  angle  c  to  be  bisected, 
or  divided  into  two  equal  parts  by  the  line 
CD,  making  the  angle  acd.  equal  to  the 
angle  bc©. 

^  Then, 


THT:0REMS.  283 

Then,  the  two  tr!3n2:les  acd,  bcd,  have  two  sides  and 
the  <:  ."t  ur;cf]  ^ru:U'  of  the  one,  equal  to  two  sides  and  the 
cop.tainefl  a^^5ie  of  the  other,  viz.  the  side  ac  equal  to  bc,  the 
BM^Jo  A.D  <^quU  to  BCD,  and  the  side  cd  common  ;  there- 
fore thcf^;  tvso  triangles  are  indentical,  or  equal  in  all  re- 
Sj)!-  /s  (th.  1)  ;  and  consequently   the   angle  a  equal  to  the 

a-VJ.     B.       Q.    E.    I). 

Corol.  1,  Honce  the  line  which  hisects  the  verticle  angle  of 
aa  isosceles  triani^lc,  bisects  the  base,  and  is  also  perpendicu- 
lar to  it. 

Corol.  2-.  Hence  too  it  appears,  that  every  equilateral  tri- 
angle, is  also  equian^uiar,  or  has  all  its  angles  equal. 

THEOREM  IV. 

WtfEN  a    Tf iangK^    has   Two  of  its  x\ngles  ^equal,  the  Sides 
Opposite  to  them  are  also  equal. 

If   the    triargle    abc,    have    the    angle   a  C 

eq'ial  to  th^  angle  b.  it  will  also  have  the  side  /K 

AC  oqijal  to  the  side  bc  /  I  \ 

For,  conceiye  tb'ii  side  ab  to  be   bisected  /    I    \ 

in   li»e    point   d,    making   ad   equal   to   db  ;  /       t      \ 

and  j.nn  DC,  dividing  the   whvol.'^  triangle  into         A.     D     B 
the  two   triangles   acd,  bcd.     Also  c  orrceive 
the  t'iangle  acd  to  be  turned  over  upon  the 
triangle  bcd,  so  that  ad  maj  f  :U  on  bd. 

Then,  because  the  line  ad  is  equal  to  the  line  db  ^by  hyp.), 
the  point  a  coincides  with  the  point  b,  and  the  point  d  with  the 
point  D.  Also,  because  the  angle  a  is  equal  to  the  an5;;le  b  by 
(hyp.),  the  hne  ac  will  fall  on  the  line  bc,  and  the  extremity  c 
of  the  side  ac  will  coincide  with  the  extremity  c  of  the  side  bc, 
becanse  dc  is  common  to  both  ;  coiisequently  the  side  ac  is 
equal  to  bc     q.  e.  d*. 

Corol.  Hence  every  eiijuiangular  triangle  is  also  equila- 
teral. 

THEOREM  V. 

When  Two  Triangles  have  all  the  Three  Sides  in  the  one, 
equal  to  all  the  Three  Sides  in  the  other,  the  Triangles  are 
Identical,  or  have  also  their  Three  Angles  equal,  each  to  each. 

Let  the  two  triangles,  abc,  abd, 
have  their  three  sides  respectively 
equal,  viz.  the  side  a«  equal  to  ab, 
AC  to  AD,  and  bc  to  bd  ;  then  shall 
the  two  triangles  be  identical, or  have 
their  angles  equal,  viz.  those  angles 

*  I  lis  tlemon-iti-ation  of  Theoiem  iv,  does  not  appear  to  me  to  be 
conclusive.     Editoii.  • 

that 


^84  GEOMETR\. 

that  are  opposite  to  the  equal  sides  ;  C 

namely,  the  angle  bag  to  the  angle  ^--^^"^^Nv 

BAD,  the  angle  abc  to  the  angle  abd,  a  <^— -__J__\-d 

and  the  angle  c  to  the  angle  d.  ^;^~—     \~/^ 

For,  conceive  the  two  triangles  to  ^^""^-j/ 

be  joined  together  by  their  longest  D 

equal  sides,  and  draw  the  line  cd. 

Then,  in  the  triangle  acd,  because  the  side  ac  is  equal 
to  AD  (by  hyp.),  the  angle  acd  is  equal  to  the  angle  adc 
(th.  3).  In  like  manner,  in  the  triangle  bcd.  the  angle  bcd  is 
equal  to  the  angle  bdc,  because  the  side  bc  is  equal  to  bd. 
Hence  then,  the  angle  acd  being  equal  to  the  angle,  a©c  and 
the  angle  bcd  to  the  angle  bdc,  by  equal  additions  the  sum  of 
the  two  angles  acd,  bcd,  is  equal  to  the  sum  of  the  two  adc, 
BDC,  (ax.  2),  that  is,  the  whole  angle  acb  equal  to  the  whole 
angle  adb. 

Since  then,  the  two  sides  ac,  cb,  are  equal  to  the  two  sides 
AD,  db,  each  to  each,  (by  hyp.),   and  their  contained  angles 
ACB,  adb,  also  equal,  the  two  triangles  abc,  and  abd,  are  iden- 
tical (th.  1),  and  have  the  other  angles  equal,  viz.  the   angle" 
BAc   to  the  angle  bad,  and  the  angle  abc  to  the  angle  abd. 

Q.    E.    D. 


THEOREM  VI. 

When  one  Lme  meets  another,  the  Angles  which  it  makes 
on  the  Same  Side  of  the  other,^  are  together  ^qual  to  Two 
Right  Angles. 

Let  the  line  ab  meet  the  line  cd  :  then 
TPill  the  two  angles  abc,  abd,  taken  together, 
be  equal  to  two  right  angles. 

For,  first,  when  the  tw®  angles  abc,  abd  . 
are  equal  to  each  other,  they  are  both  of 
them  right  angles  (def.  15.) 


E     A 


B       D 

But  when  the  angles  are  unequal,  suppose  be  drawn  per- 
pendicular to  CD.  Then,  since  the  two  angles  ebc,  ebd,  are 
right  angles  (def.  15),  and  the  angle  ebd,  is  equal  to  the  two 
angles  eba,  abd,  together  (ax.  8),  the  three  angles,  ebc,  eba, 
and  abd,  are  equal  to  two  right  Angles. 

But  the  two  angles  ebc,  eba,  are  together  equal  to  the 
angle  abc  (ax.  8).  Consequently  the  two  angles  abc,  abd,  are 
also  equal  to  two  right  angles,     q.  e.  d. 

Corol.  1.  Hence  also,  conversely,  if  the  two  angles  abc, 
abd,  on  both  sides  of  the  line  ab,  make  up  together  two 
right  angles,  then  cb  and  bd  form  one  continued  right  line  cd. 

Corol, 


THEOREMS.  286 

Corol.  2.  Hence,  all  the  angles  which  can  be  made,  at 
any  point  b,  by  any  number  of  lines,  on  the  same  side  of 
the  right  line  cd,  are,  when  taken  all  together,  equal  to  two 
right  angles. 

Corol.  3.  And,  as  all  the  angles  that  can  be  made  on  the 
other  side  of  the  line  cd  are  also  equal  to  two  right  angles  ; 
therefore  ail  the  angles  that  can  be  made  quite  round  a  point 
B,  by  any  number  of  lines,  are  equal  to  four  right  angles. 

Corol.  4.  Hence  also  the  whole  circumfer- 
ence of  a  circle,  being  the  sum  of  the  mea- 
sures of  all  the  angles  that  can  be  made  about 
the  centre  f  (def.  67),  is  the  measure  of  four 
right  angles.  Consequently ,^  a  semicircle,  or 
180  degrees,  is  the  measure  of  two  right 
angles  :  and  a  quadrant,  or  90  degrees,  the  measure  of  one 
right  angle. 

THEOREM  Vn. 

When  two  Lines  Intersect  each  other,  the  Opposite  Angles 
are  equal. 

Let  the  two  lines  ab,  cd,  intersect  in 
the  point  e  ;  then  will  the  angle  aec  be 
equal  to  the  angle  bed,  and  the  angle 
AED  equal  to  the  angle  ceb. 

For  since  the  Hne  ce  meets  the  line 
AB,  the  two  angles  aec,  bec,  taken  to- 
gether, are  equal  to  two  right  angles  (th.  6). 

In  like  manner,  the  line  be,  meeting  the  line  cd,  makes 
the  two  angles  bec,  bed,  equal  to  two  right  angles. 

Therefore  the  sum  of  the  two  angles  aec,  bep,  is  equal 
to  the  sum  of  the  two  bec,  bed  (ax.  1). 

And  if  the  angle  bec,  which  is  common,  be  taken  away 
from  both  these,  the  remaining  angle  aec  will  be  equal  to 
the  remaining  angle  bed  (ax.  3). 

And  in  like  manner  it  may  be  shown,  that  the  angle  aed 
is  equal  to  the  opposite  angle  bec. 


THEOREM  Vm. 

When  One  Side  of  a  Triangle  is  produced,  the  Outward 
Angle  is  Greater  than  either  of  the  two  Inwajrd  Opposite 
Angles. 

Let 


286 


GEOMETRY. 


Let  ABC  be  a  triangle,  having  the 
side  AB  produced  to  d  ;  then  will  the 
outward  angle  cbd  be  greater  than 
either  of  the  inward  opposite  angles  a 
or  c. 

For,  conceive  the  side  'bc  to  be  bi- 
sected in  the  point  e,  and  drnw  the  line 
AE,  producing  it  till  ef  be  equal  to  ae  ; 
and  join  BF.  - 

Then,  since  the  two  triangles  aec,  bef,  have  the  side 
ae  =  the  side  ef,  and  the  side  ce  =  the  side  be  (by  suppos.) 
and  the  inrJuded  or  opposite  angles  at  e  also  equal  (tb.  7), 
therefore  those  t;Vo  triangles  are  equal  in  all  respects 
(th.  1),  and  have  the  angle  c«=  tht  corresponding  angle  ebf. 
Bat  the  angle  cbd  is  greater  tha»  the  angle  kbf  ;  consequent- 
ly the  said  outward  angle  cbd  is  also  great^T  than  the  angle  c. 

In  like  manner,  if  cb  be  prod-jced  to  o.  and  ab  be  bisected, 
it  Diaj  be  shown  that  the  outward  angle  abg,  or  its  equal  cbd, 
is  greater  than  the  other. angle  a. 


THEOREM  IX. 


DB 


') 


The  Greater  Side,  of  every  Triangle,  is  opposite  to  the 
Greater  Angle  ;  and  the  Greater  Angle  opposite  to  the  Great- 
er Side. 

Let  ABC  be  a  triangle,  having  the  side 
ab  greater  than  the  side  ac  ;  then  will 
the  angle  acb.  opposite  the  greater  side 
AB,  !>e  greater  than  the  angle  b,  opposite 
the  less  side  ac^ 

For,  on  the  greater  side  ab,  take  the 
part  ar  equal  to  thS  less  side  ac,  and  join  cd.  Then,  sin(ie 
BCD  is  a  triangle,  the  outward  angle  adc  is  greater  than  the 
inward  oppoj^ite  angle  b  (th.  8).  But  the  angle  acd  is  equal 
to  .tbe  said  outward  angle  adc,  because  ad  is  equal  to  ac  (th.  3). 
Consequently  the  asigle'  acd  also  is  greater  than  the  angle  b. 
And  siace  the  angle  acd  is  only  a  part  of  acb,  much  more 
must  the  whole  angle  acb  be  greater  than  the  angle  b.    q.  e.  d. 

Again,  conversely,  if  the  angle  c  be  greater  than  the  angle 
B,  then  will  the  side  ab,  opposite  the  former,  be  greater  than 
the  side  ac,  opposite  the  latter. 

For,  if  ab  be  not  greater  than  ac,  it  must  be  either  equal 
to  it,    or  less  than  it.     But   it  cannot   be   equal,   for  then 


THEOREMS.  287 

the  angle  c  would  be  equal  to  the  angle  b  (th.  3)  which  it  is 
not,  by  the  supposition.  Neither  can  it  be  less,  for  then  the 
auji^le  c  would  be  le!?s  than  the  angle  b,  by  the  former  part  of 
this;  ;  which  is  also  contrary  to  the  supposition.  The  side  ab, 
then,  being  neiilier  equal  to  ac,  nor  less  than  it,  must  neces- 
sarily be  greater,     q.  e.  d. 

THEOREM  X. 

The   Sum  of  any  Two   Sides  of  a  Triangle  is  Greater  than 
the  Third  Side.  ^ 

Let  ABC  be  a  triano;le  ;  then  will  the 
sum  of  any  two  of  its  sides  be  greater  than 
the  third  side,  as  for  instances,  ac  +  cb 
greater  than  as. 

For,  pi'oduce  ac  till  cp  be  equal  to 
CB,  or  AD  equal  to  the  sum  of  the  two 
AC  +  CB  ;  and  join  bd  : — Then,  because  ^  ~~B 

CD  is  equal  to  cb  (by  «onstr.),  the  angle  d  is  equal  to  the  angle 
CBD  (th.  3).  But  the  angle  abd  is  greater  than  the  angle  cbd, 
consequently  it  must  also  be  greater  than  the  angle  d.  And, 
since  the  greater  side  of  any  triangle  is  opposite  to  the' greater 
angle  (th.  9),  the  side  ad  (of  the  triangle  abd)  is  greater  than 
the  side  ab.  "  But  ad.  is  equal  to  ac  and  cd,  or  ac  and  cb, 
taken  together  (by  constr.)  ;  therefore  ac  -{-  cb  is  also  great- 
er than  AB.     Q.  E.  d. 

CoroL  The  shortest  distance  between  two  points,  is  a  single 
right  line  drawn  from  the  one  point  to  the  other. 

THEOREM  XI. 

The  Difference  of  any  Two  Sides  of  a  Triangle,  is  Less  than 
the  Third  Side. 

Let  ABC  be  a  triangle  ;  then  will  the 
difference  of  any  two  sides,  as  ab  —  ac, 
be  less  than  the  third  side  bc. 

For,  produce  the  less  side  ac  to  d,  till 
ad  be  equal  to  the  greater  side  ab,  so 
that  CD  may  be  the  difference  of  the  two 
sides  AB  —  AC  ;  and  join  bd.  Then,  be- 
cause AD  is  equal  to  ab  (by  constr.),  the  opposite  angles  o 
and  ABD  are  equal  (th.  3).  But  the  angle  cbd  is  less  than  the 
angle  abd,  and  consequently  also  less  than  the  equal  angle  d. 
And  since  the  greater  side  of  any  triangle  is  opposite  to  the 

greater 


2*8  GEOMETRY. 

greater  angle  (th.  9),  the  side  cd  (of  the  triangle  bcd)  is  less 
than  the  side  bc.     q..  e.  d. 

THEOREM  Xn. 

When  a  Line   Intersects  two  Parallel  Lines,   it  makes  the 
Alternate  Angles  Equal  to  each  other. 

Let  the  line  ef  cut  the  two  parallel 
lines  AB,  CD  ;  then  will  the  angle  aef  be 
equal  to  the  alternate  angle  efd. 

For  if  they  are  not  equal,  one  of  them 
must  be  greater  than  the  other  ;  let  it  be 
EFD  for  instance  which  is  the  greater,  if 
possible  ;  and  conceive  the  line  fb  to  be 
drawn  ;  cutting  off  the  part  or  angle  efb  equal  to  the  angle 
AEF  ;  and  meeting  the  line  ab  in  the  point  b. 

Then,  since  the  outward  angle  aef,  of  the  triangle  bef,  is 
greater  than  the  inward  opposite  angle  liFB  (th.  8)  ;  and  since 
these  two  angles  also  are  equal  (by  the  constr.)  it  follows,  that 
those  angles  are  both  equal  and  unequal  at  the  same  time  : 
which  is  impossible.  Therefore  the  angle  efd  is  not  unequal 
to  the  alternate  angle  aef,  that  is,  they  are  equal  to  each 
other.     Q.  e.  d. 

Corol.  Right  lines  which  are  perpendicular  to  one,  of  tw« 
parallel  lines,  are  also  perpendicular  to  the  other. 

THEOREM  Xin. 

When  a  line,  cutting  Two  other  Lines,  makes  the  Al- 
ternate Angles  Equal  to  each  other,  those  two  Lines  are  Pa- 
rallel. 

Let  the  line  ef,  cutting  the  two  lines 
ab,  CD,  make  the  alternate  angles  aef, 
»fe,  equal  to  each  other  ;  then  will  ab 
be  parallel  to  cd. 

For  if  they  be  not  parallel,  let  some  -     ^ _ 

•ther  line,  as  fg,  be  parallel  to  ab.  ^  /lE"  '""---P 
Then,  because  of  these  parallels,  the  ^ 

angle  aef  is  equal  to  the  alternate  angle  efg  {th.  12,  But 
\he  angle  aef  is  equal  to  the  angle  efd  (by  hyp.).  There- 
fore the  angle  efd  is  equal  to  the  angle  efg  (ax.  1)  ;  that  is, 
a  part  is  equal  to  the  whole,  which  is  impossible.  Therefore 
BO  line  but  cd  can  be  parallel  to  ab.     q.  e.  d. 

Corol.  Those  Hues  which  are  perpendicular  to  the  same 
line,  are  parallel  to  each  other. 

THEOREM 


THEOREMS,  28^ 

THEOREM  XrV. 

When  a  Line  cuts  two  Parallel  Lines,  the  Outward  Angle 
is  Equal  to  the  Inward  Opposite  one,  on  the  Same  Side  ;  and 
the  two  inward  Angles,  on  the  Same  Side,  equal  to  two  Right 
Angles. 

Let  the  line  ef  cut  the  two  parallel 
lines  AB,  CD  ;  then  will  thei)utward  angle 
EGB  be  equal  to  the  inward  opposite  an- 
gle GHD,  on  the  same  side  of  the  line  ef; 
and  the  two  inward  angles  bgh,  ghd, 
taken  together,  will  be  equal  to  two 
right  angles. 

For,  since  the  two  lines  ab,  cd,  are 
parallel,  the  angle  agh  is  equal  to  the  alternate  angle  ghd,. 
(th.   12).     But  the  angle  agh  is  equal  to  the  opposite  angle 
EGB  (th.   7).     Therefore  the  angle  egb  is  also  equal  to  the 
angle  ghd  (ax.  1)-.  q.  e.  d. 

Again,  because  the  two  adjacent  angles  egb,  bgh,  are  to- 
gether equal  to  two  right  angles  (th.  6)  ;  of  which  the  an- 
.  gle  egb  has  been  shown  to  be  equal  to  the  angle  ghd  ;  there- 
fore the  two  angles  bgh,  ghd,  taken  together,  are  also  equal 
to  two  right  angles. 

Corol.  1.  And,  conversely,  if  one  line  meeting  two  other 
lines,  make  the  angles  on  the  same  side  of  it  equal,  those 
two  lines  are  parallels. 

Corol.  2.  If  a  line,  cutting  two  other  lines,  make  the  sum 
©f  the  two  inward  angles,  on  the  same  side,  less  than  two 
right  angles,  those  two  lines  will  not  be  parallel,  but  will  m^et 
each  other  when  produced. 

THEOREM  XV. 

Those  Lines  which  are  Parallel  to  the  Same  Line,  are  Pa- 
rallel to  each  other. 

Let  the   Lines  ab,  cd,  be  each  of  -h 

them  parallel  to  the  line  ef  ;  then  shall      J^ 1 ^g 

the  lines  ab,  cd,  be  parallel  to  each      ^ | -p. 

other.  ^        Hi 

For,  let  the  line  gi  be  perpendicular      £  X  F 

to  ef.     Then  will  this  line  be  also  per- 
pendicular to  both  the  hnes  ab,  cd  (corol    th.  12),  and  con- 
sequently the  two  lines  ab,  cd,  are  parallels  (corol.  th.  13). 

Q.    E.    D. 

Vol.  L  38  THEOKBM 


290  GEOMETRY. 


THEOREM  XVL 

When  one  Side  of  a  triangle  is  produced,  the  Outward 
Angle  is  equal  to  both  the  inward  Opposite  Angles  taken  to- 
gether. 

Let  the  side,  ab,  of  the  triangle                   C              J| 
ABC,  be  produced  to  d  ;  then  will  the                   /\             / 
outward  angle  cbd  be  equal  to  the               /     \       / 
sum  of  the  two  inward  opposite  an-             /       \'/ 
gles  A  and  c.  ^ g 5 

For,  conceive  be  to  be  drawn  pa- 
rallel  to  the  side  ac  of  the  triangle. 

Then  bc,  meeting  the  two  parallels  ac,  be,  'makes  the  alter- 
nate angles  c  and  cbe  equal  (th.  12).  And  ad,  cutting  the 
same  two  parallels  ac,  be,  makes  the  inward  and  outward 
angles  on  the  same  side,  a  and  ebd,  equal  to  each  other  (th. 
14).  Therefore,  by  equal  additions,  the  sum  of  the  two  an- 
gles A  and  c,  is  equal  to  the  sum  of  the  two  cbe  and  ebd,  that 
is,  to  the  whole  angle  cbd  (by  ax.  2).  «i.  e.  d. 

THEOREM  XVII. 

In  any  Triangle,  the  sum  of  all  the  Three  Angles  is  equal  t© 
Two  Right  Angles. 


C 

A. 


Let  ABC  be  any  plane  triangle  ;  then 
the  sum  of  the  three  angles  a  -f"  b  "h  c 
is  equal  to  two  right  angles. 

For,  let  the  side  ab  be  produced  to  d. 

Then  the  outward  angle   cbd  is   equal         ^ ■ — ^ *v 

to  the  sum  of  the  two  inward  opposite 
angles  a  -f  c  (th.  16).  To  each  of  these  equals  add  the 
inward  angle  b,  then  will  the  sum  of  the  three  inward  an- 
gles A  -f  B  -f  c  be  equal  to  the^um  of  the  two  adjacent  angles 
ABC  -{-  CBD  (ax.  2).  But  the  sum  of  these  two  last  adjacent 
angles  is  equal  to  two  right  angles  (th.  6).  Therefore  also 
the  sum  of  the  three  angles  of  the  triangle  a  -f  b  -|-  c  ia 
equal  to  two  right  angles  (ax.  1).  q.  e.  d. 

Carol.  1 .  If  two  angles  in  one  triangle,  be  equal  to  two 
angles  in  another  triangle,  the  third  angles  will  also  be  equal 
(ax.  3),  and  the  two  triangles  equiangular. 

Cbrol.  2.  If  one  angle  in  one  triangle  be  equal  to  one 
angle  in  another,  the  sums  of  the  remaining  angles  will  also 
be  equal  (ax.  3.) 


THEOREMS. 


251 


Corol.  3.  If  one  angle  of  a  triangle  be  richt,  the  sum  of 
the  other  two  will  also  be  equal  to  a  right  ani^le,  and  each 
of  them  singly  will  be  acute,  or  less  than  a  right  angle. 

Corol.  4.  The  two  least  angles  of  every  triangle  are  acute, 
or  each  less  than  a  right  angle. 


l^HEOREM  XVin. 

In  any  Quadrangle,  the  sum  of  all  the  Four  Inward  Angles, 
is  equal  to  Four  Right  Angles. 

Let  ABCD  be  a  quadrangle  ;  then  the 
sum  of  the  four  inward  angles,  a  +  »  4* 
c  -j-  D  is  equal  to  four  ri^ht  angles. 

Let  the  diagonal  ac  be  drawn,  dividing 
the  quadrangle  into  two  triangles,  ABC,  ADC. 
Then,  because  the  sum  of  the  three  angles 
of  each  of  these  triangles  is  equal  to  two  A  B 

right  angles  (th.  17)  ;  it  follows,  that  the  sum  of  all  the 
angles  of  both  triangles,  which  make  up  the  four  angles  of 
the  quadrangle,  must  be  equal  to  four  right  angles  (ax.  2). 

Q.   E.  D. 

Corol.  1.  Hence,  if  three  of  the  angles  be  right  ones,  the 
fourth  will  also  be  a  right  angle. 

Corol.  2.  And,  if  the  sum  of  two  of  the  four  angles  be 
equal  to  two  right  angles,  the  sum  of  the  remaining  two  will 
also  be  equal  to  two  right  angles. 


THEOREM  XIX. 

In  any  figure  whatever,  the  Sum  of  all  the  Inward  Angles, 
taken  together,  is  equal  to  Twice  as  many  Right  Angles, 
wanting  four,  as  the  Figure  has  Sides 

Let  ABCDE  be  any  figure  ;  then  the 
sum  of  all  its  inward  angles,  a  -f  b  + 
c  -}-  D  -|-  E,  is  equal  to  twice  as  many 
right  angles,  wanting  four,  as  the  figure 
has  sides. 

For,  from  any  point  p,  within  it,  draw 
lines  PA,  PB,  PC,  kc.  to  all  the  angles, 
dividing  the  polygon  into  as  many  tri- 
angles as  it  has  sides.  Now  the  sum  of  the  three  angles  of 
each  of  these  triaugles,  is  equal  to  two  right  angles  (th.  17)  ; 
therefore  the  sum  of  the  angles  of  all  the  triangles  is  equal 
to  Twice  as  miny  right  angles  as  the  figure  has  sides.  But 
the  sum  of  .all  the  angles  about  the  point  p,  which  are  so 

manj 


29f 


GEOMETRY. 


many  of  the  angles  of  the  triangles,  but  no  part  of  the  in- 
ward angles  of  the  polygon,  is  equal  to  four  right  angles 
(cord.  3,  th  6).  and  must  be  deducted  out  of  the  former 
sum.  Hence  it  follows  that  the  sum  of  all  the  inward  angles 
of  the  polygon  alone,  A-j-B-f'C-l-D-f'E,  is  equal  to  twice 
as  many  right  angles  as  the  figure  has  sides,  wanting  the 
said  four  right  angles.  Q.  e.  d, 

THEOREM  XX. 

When  every  Side  of  any  Figure  is  produced  out,  the 
Sum  of  all  the  Outward  Angles  thereby  made,  is  equal  to 
Four  Right  Angles. 

Let  A,  B,  c,  &c.  be  the  outward 
angles  of  any  polygon,  made  by  pro- 
ducing all  the  sides  ;  then  will  the  sum 
A  -f  B  +  c  +  D  +  E,  of  all  those  outward 
angles,  be  equal  to  four  right  angles. 

For  every  one  of  these  outward  angles, 
together  with  its  adjacent  inward  angle, 
make  up  two  right  angles,  as  a  -f  a  equal 
to  two  right  angles,  being  the  two  angles 
made  by  one  line  meeting  another  (th.  6).  And  there 
being  as  many  outward,  or  inward  angles,  as  the  figure  has 
sides  :  therefore  the  sum  of  all  the  inward  and  outward 
angles,  is  equal  to  twice  as  many  right  angles  as  the  figure 
has  sides.  But  the  sum  of  all  the  inward  angles,  with  four 
right  angles,  is  equal  to  twice  as  many  right  angles  as  the 
figure  has  sides  (th.  19).  Therefore  the  sum  of  all  the  in- 
ward and  all  the  outward  angles,  is  equal  to  the  sum  of  all 
the  inward  angles  and  four  right  angles  (by  ax.  1).  From 
each  of  these  take  away  all  the  inward  angles,  and  there 
remain  all  the  outward  angles  equal  to  four  right  angles 
(by  ax.  3). 

* 

THEOREM  XXI. 

A  Perpendicular  is  the  Shortest  Line  that  can  be  drawn 
from  a  Given  Point  to  an  Indefinite  Line.'  And,  of  any- 
other  Lines  drawn  from  the  same  Point,  those  that  are  Nearest 
he  Perpendicular,  are  Less  than  those  More  Remote. 

If  AB,  AC,  ao,  &c.  be  lines  drawn  from 
the  given  point  a,  to  the  indefinite  line  de, 
of  which  AB  is  perpendicular.  Then  shall 
the  perpendicular  ab  be  less  than  ac,  and 
AC,  less  than  ad,  &;c. 

For,  the  angle  b  being  a  right  one,  the 


THEOREMS.  293 

angle  c  is  acute  (by  cor.  3,  th.  17),  and  therefore  less  than 
the  angle  b.  But  the  less  angle  of  a  triangle  is  subtended  by 
the  less  side  (th.  9).  Therefore  the  side  ab  is  less  than  the 
side  AC. 

Again,  the  angle  acb  being  acute,  as  before,  the  adjacent 
angle  acd  will  be  obtuse  (by  th.  6)  ;  consequently  the  angl« 
»  is  acute  (corol.  3,  th.  17), and  therefore  is  less  than  the  angle 
€.  And  since  the  less  side  is  opposite  to  the  less  angle,  there- 
fore the  side  ac  is  less  than  the  side  ad.  -  q.  e.  d 

Corol.  A  perpendicular  is  the  least  distant  of  a  given  point 
^om  a  line. 


THEOREM  XXII. 

The  Opposite  Sides  and  Angles  of  any  Parallelogram  are 
equal  to  each  other ;  and  the  Diagonal  divides  it  into  two 
Equal  Triangles. 

Let  abcd  be  a  parallelogram,  of  which  ^^  P 

the  diagonal  is  bd  ;  then  will  its  opposite 
sides  and  angles  be  equal  to  each  other, 
and  the  diagonal  bd  will  divide  it  into  two 
equal  parts,  or  triangles. 

For,  since  the  sides  ae  and  do  are  pa- 
rallel, as  also  the  sides  ad  and  bc  (detin. 
32),  and  the  line  bd  meets  them  ;  therefore  the  alternate 
angles  are  equal  (th.  12),  namely,  the  angle  aed  to  the  angle 
€db,  and  the  angle  adb  to  the  angle  cbd.  Hence  the  two 
triangles,  having  two  angles  in  the  one  equal  to  two  angles 
ia  the  other,  have  also  their  third  angles  equal  (cor:  1,  th.  17), 
namely,  the  angle  a  equal  to  the  angle  c,  which  are  two  of  the 
opposite  angles  of  the  parallelogram. 

Also,  if  to  the  equal  angles  abd,  cdb,  be  added  the  equal 
angles  cbd,  adb,  the  wholes  will  be  equal  (ax.  2),  namely,  the 
whole  angle  abc  to  the  whole  ado,  which  are  the  other  two 
opposite  angles  of  the  parallelogram.  q.'  e.  d. 

Again,  since  the  two  triangles  are  mutually  equiangular  and 
have  a  side  in  each  equal,  viz.  the  common  side  bd  ;  there- 
fore the  two  triangles  are  identical  (th.  2),  or  equal  in  all  re- 
spects, namely,  the  side  ab  equal  to  the  opposite  side  dc,  and 
ad  equal  to  the  opposite  side  bc,  and  the  whole  triangle  abd 
equal  to  the  whole  triangle  BCB.  q.  e.  d, 

Corol. 


294  GEOMETRY. 

Corol.  1.  Hence,  if  one  angle  of  a  parallelogrskm  be  a  right 
angle,  all  the  other  three  will  also  be  li^-ht  angles,  and  the 
parallelogram  a  rectangle. 

Corol.  2.  Hence  also,  the  sam  of  any  two  adjacent  angles  of 
a  parallelogram  is  equal  to  two  right  angles. 

THEOREM  XXIII. 

Every  Quadrilateral,  whose  Opposite  Sides  are  equal,  is  a 
Parallelogram,  or  has  its  Opposite  Sides  Parallel. 
Let  ABCD  be  a  quadrangle  having  the 
opposite  sides  equal,  namely,  the  side  ab 
equal  to  dc,  and  ad  equal  to  eg  ;  then 
shall  these  equal  sides  be  also  parallel, 
and  the  figure  a  parallelogram. 

For,  let  the  diagonal    bd  be  drawn. 
Then,   the    triangles,    abd,    cbd,    being 
mutually  equilateral  (by  hyp.),  they  are 
also  mutually  equiangular  (th.  6),  or  have  their  correspond- 
ing angles  equal  ;  consequently  the  opposite  sides  are  parallel 
(th.  13)  ;   viz.  the  side  ab  parallel  to  do,  and  ad  parallel  to  bc, 
and  the  figure  is  a  parallelogram.  q.  e    d. 

THEOREM  XXIV. 

Those  Lines  which  join  the  Corresponding  Extremes  of 
two  Equal  and  Parallel  Lines,  are  themselves  Equal  and 
Parallel.  ^^ 

Let  ab,  DC,  be  two  equal  and  parallel  lines  ;  then  will  the 
lines  AD,  bc,  which  join  their  extremes,  be  also  equal  and  pa- 
rallel.    [See  the  fig.  above.] 

For,  draw  the  diagonal  bd.  Then,  because  ab  and  dc  are 
parallel  (bv  hyp),  the  angle  abd  is  equal  to  the  alternate 
angle  bdc  (th  12),  Hence  then,  the  two  triangles  having  two 
sides  and  the  contained  angles  equal,  viz.  the  side  ab  equal  to 
the  side  dc,  and  the  side  bd  common,  and  the  contained 
angle  abd  equal  to  the  contained  angle  bdc,  they  have  the 
remaining  sides  and  angles  also  respectively  equal  (th.  1)  ; 
consequently  ad  is  equal  to  bc,  and  also  parallel  to  it  (th.  12). 

Q.    E.    D. 
THEOREM  XXV. 

Parallelograms,  as  also  Triangles,  standing  on  the  Same 
Base,  and  between  the  Same  Parallels,  are  equal  to  each 
«ther. 

Let 


THEOREMS, 


id's 


Let  ABCD,  ABEF,  be  two  parallelo- 
grams, and  ABC,  ABF,  two  triangles, 
stalling  on  the  same  base  ab,  and  be- 
tween the  same  parallels  ab,  de  :  then 
will  the  parallelogram  ABCD,be  equal  to 
the  parallelogram  ABEF,  and  the  triangle 
ABC  equal  to  the  triangle  abf. 

For,  since  the  line  de  cuts  the  two 
parallels  ap,  be,  and  the  two  ad,  bc,  it  makes  the  angle  e 
equal  to  the  angle  afd,  and  the  angle  d  equal  to  the  angle  bce 
(th.  14)  ;  the  two  triangles  adf,  bce,  are  therefore  equiangu- 
lar (cor.  1,  th.  17)  ;  and  having  the  two  corresponding  sides, 
ad,  bc,  equal  (th.  22),  being  opposite  sides  of  a  parallelogram, 
these  two  triangles  are  identical,  or  equal  in  all  respects 
(th.  2).  If  each  of  these  equal  triangles  then  be  taken  from 
the  whole  space  abed,  there  will  remain  the  parallelogram 
ABEF  in  the  one  case,  equal  to  the  parallelogram  abcd  in  the 
other  (by  ax.  3). 

Also  the  triangles  abc,  abf,  on  the  same  base  ab,  and  be- 
tween the  same  parallels,  are  equal,  being  the  halves  of  thfe 
said  equal  parallelograms  (th.  22).  q.  E   d. 

Carol.  1.  Parallelograms,  or  triangles,  having  the  same  base 
and  altitude,  are  equal.  For  the  altitude  is  the  same  as  the 
perpendicular  or  distance  between  the  two  parallels,  which  is 
every  where  equal,  by  the  definition  of  parallels. 

Carol.  2.  Parallelograms,  or  triangles,  having  equal  bases 
and  altitudes,  are  equal.  For,  if  the  one  figure  be  applied 
with  its  base  on  the  other,  the  bases  will  coincide  or  be  the 
same  because  they  are  equal  :  and  so  the  two  figures,  having 
the  same  base  and  altitude,  are  equal. 


THEOREM  XXVI. 


3)        C      E 


If  a  Parallelogram  and  a  Triangle  stand  on  the  Same  Base, 
and  between  the  Same  Parallels,  the  Parallelogram  will  be 
Double  the  Triangle,  or  the  Triangle  Half  the  Parallelogram. 

Let  abcd  be  a  parallelogram,  and  abe,  a 
triangle,  on  the  same  base  ab,  and  between 
the  same  parallels,  ab,  de  ;  then  will  the  pa- 
rallelogram ABCD  be  double  the  triangle  abe, 
or  the  triangle  half  the  parallelogram. 

For,  draw  the  diagonal  ac  of  the  parallelo- 
gram, dividing  it  into  two  equal  parts  (th.  22). 
Tken  because  the  triangles  abc,  abe,  on  the 

same 


296  GEOMETRY. 

same  base,  and  betfveen  the  same  parallels,  are  equal  (th.  25)  ; 
and  because  the  one  triangle  abc  is  half  the  parallelogram 
ABCD  (th.  22),  the  other  equal  triangle  abe  is  also  equal  to  half 
the  same  parallelogram  abcd.  q.  e.  d. 

Corol.  1.  A  triangle  is  equal  to  half  a  parallelogram  of  the 
same  base  and  altitude,  because  the  aHitude  is  the  perpendi- 
cular distance  between  the  parallels,  which  is  every  where 
equal,  by  the  definition  of  parallels. 

Corol.  2.  If  the  base  of  a  parallelogram  be  half  that  of  a 
triangle,  of  the  same  altitude,  or  the  base  of  the  triande  be 
double  that  of  the  parallelogram,  the  two  figures  will  be  equal 
to  each  other. 

THEOREM  XXVn. 

Rectangles  that  are  contained  by  Equal  Lines,  are  Equal 
to  each  other. 

Let  BD,  FH,  be  two  rectangles,  having        J) C'  H     G 

the  sides  ab,  bc,  equal  to  the  sides  ef,  fg, 
each  to  each  ;  then  will  the  rectangle  bd 
be  equal  to  the  rectangle  fh. 


/ 

For,  draw  the  two  diagonals  ac,  eg,  di-  j^  j^  j^  ]gi 
Tiding  the  two  parallelograms  each  into 
two  equal  parts.  Then  the  two  triangles 
ABC,  efg,  are  equal  to  each  other  (th.  1),  because  they  have 
the  two  sides  ab,  bc,  and  the  contained  angle,  b  equal  to  the 
two  sides  ef,  fg,  and  the  contained  angle  p  (by  hyp).  But 
these  equal  triangles  are  the  halves,  of  the  respective  rectan- 
angles.  And  because  the  halves,  or  the  triangles,  are  equal, 
the  wholes,  or  the  rectangles,  db,  hf,  are  also  equal  (by  ax.  6). 

Q.  E.  i>. 

Carol.     The  squares  on  equal  lines  are  also  equal  ;  for 
every  square  is  a  species  of  rectangle. 

THEOREM  XXVUI. 

The  Complements  of  the  Parallelograms,  which  are  about 
the  Diagonal  of  any  Parallelogram,  are  equal  to  each  other. 

Let  AC  be  a  parallelogram,  bd  a  dia- 
gonal, eif  parallel  to  ab  or  dc,  and  gih 
parallel  to  ad  or  bc,  making  ai,  ic  com- 
plements to  the  parallelograms  eg,  hf, 
which  are  about  the  diagonal  db  :  then  ^  jj 

will  the  complement  ai  be  equal  to  the 
complement  IC. 

For, 


THEOREMS. 


297 


For,  since  the  diagonal  db  biserts  the  three  parallelograms 
AC,  EG,  HF,  (th.  22)  ;  therefore,  the  whole  triangle  dab 
being  equal  to  the  whole  triangle  dcb,  and  the  parts  dei, 
IHB,  respectively  equal  to  the  parts  dgi,  ifb,  the  remaining 
parts  Ai,  ic,  must  also  be  equal  (by  ax.  3).  q.  e.  d. 

THEOREM  XXIX. 

A  Trapezoid,  or  Trapezium  having  two 'Sides  Parallel, 
is  equal  to  Half  a  Parallelogram,  whose  Base  is  the  Sum  of 
those  two  Sides,  and  its  Altitude  the  Perpendicular  Distance 
between  them. 


D     C  H     F 


A.    GB    E 


Let  ABCD  be  the  trapezoid,  having  its 
two  sides  ab,  do,  parallel  ;  and  in  ab 
produced  take  be  equal  to  do,  so  that 
AE  may  be  the  sum  of  the  two  parallel 
sides  ;  produce  dc  also,  and  let  ef,  gc, 
bh,  be  all  three  parallel  to  ad.  Then  is 
AF  a  parallelogram  of  the  same  altitude  with  the  trapezoid 
ABCD,  having  its  base  ae  equal  to  the  sum  of  the  parallel  sides 
of  the  trapezoid  ;  and  it  is  to  be  proved  that  the  trapezoid 
ABCD  is  equal  to  half  the  parallelogram  af. 

Now,  since  triangles,  or  parallelograms,  of  equal  bases 
and  altitude,  are  equal  (corol.  2,  th.  25),  the  parallelogram 
DG  is  equal  to  the  parallelogram  he,  and  the  triangle  cgb 
equal  to  the  triangle  chb  ;  consequently  the  line  bc  bisects, 
or  equally  divides,  the  parallelogram  af,  and  abcd  is  the 
half  of  it.  (i.  E.  D. 

THEOREM  XXX. 

The  Sum  of  .all  the  Rectangles  contained  under  one 
Whole  Line,  and  the  several  Parts  of  another  Line,  any  way 
divided,  is  Equal  to  the  Rectangle  contained  under  the  Two 
Whole  Lines. 

Let'  AD   be    the  one  line,  and   ab   the 
other,    divided    into    the    parts    ae,     ef,      ^       G  H  C 
fb  ;  then  will  the  rectangle  contained  by 
ad  and   ab,  be  equal  to   the  sum   of  the 

rectangles  of  ab   and  ae,   and  ad  and  ef,  

and  ad  and  fb   :  thus  expressed,  ad  .  ab     A.       E    P  IB 

=  AD  .  AE  -j-  AD  .  ef  +  AD  .  FB. 

For,  make  the  rectangle  ac  of  the  two  whole  lines  ad, 
AB  ;  and  draw  eg,  fh,  perpendicular  to  ab,  or  parallel  to 
AD,  to  which  they  are  equal  (th.  22).  Then  the  whole 
rectangle  ac  is    made   up    of  all  the   other  rectangles  ag, 

Vol.  L  39  eh, 


GEOMETRY. 


i>    one 


E    FB 


I  r  B 


EH,  Fc.  But  these  rectangles  are  contain- 
ed by  AD  and  ae,  eg  and  ef,  fh  and  fb  ; 
which  are  equal  to  the  rectangles  of  ad 
and  AE,  AD  and  ef,  ad  and  fb.  because 
AD  is  equal  to  ea<^h  of  the  two  eg;  fh. 
Therefore  the  rectangle  ad  .  ab  is'  equal  to  the  sunn  of  all 
the  other  rectangles  ad  .  ae,  ad  .  ef,  ad  .  fb.  q,.  e.   d. 

Carol.  If  a  right  line  be  divided  into  any  two  parts  ;  the 
square  on  the  whole  line,  is  equal  to  both  the  rectangles  of 
the  whole  line  and  each  of  the  parts. 

THEOREM  XXXI. 

The  Square  of  the  Sura  of  two  Lines  is  greater  than  the 
Sum  of  their  Squares,  by  Twice  the  Rectangle  of  the  said 
Lines.  Or,  the  Square  of  a  whole  Line,  is  equal  to  the 
Squares  of  its  two  Parts,  together  with  Twice  the  Rectangle 
of  those  Parts.  ~ 

Let  the  line  ab  be  the  sum  of  any  two 
lines  AC,  cb  :  then  will  the  square  of  ab 
be  equal  to  the  squares  of  ac,  cb,  together 
with  twice  the  rectangle  of  ac  .  cb.     That 

is,  AB2    =  AC2   -J-.CB2   -f  2aC   .  CB. 

For,  let  ABDE  be  the  square  on  the  sum  a        C    B 

or  whole  line  ab,  and   acfg   the    square 
on  the  part  ac     Produce  cf  and  gf  to   the  other  side  at  n 
and  i. 

From  the  lines  ch,  gi,  which  are  equal,  being  each 
equal  to  the  sides  of  the  square  ab  or  bd  (th.  22),  take  the 
parts  cf,  gf,  which  are  also  equal,  being  the  sides  of  the* 
square  af,  and  there  remains  fh  equal  to  fi,  which  are 
also  equal  to  dh,  di,  being  the  opposite  sides  of  a  parallelo- 
gram. Hence  the  figure  hi  is  equilateral  :  and  it  has  all 
its  angles  right  ones  (coroL  1,  th.  22)  ;  it  is  therefore  a 
square  on  the  line  fi,  or  the  square  of  its  equal  cb.  Also 
the  figures  ef,  fb,  are  equal  to  two  rectangles  under  ac 
and  cb,  because  gf  is  equal  to  ac,  and  fh  or  fi  equal 
to  cB.  But  the  whole  square  ad  is  made  up  of  the  four 
figures,  viz.  the  two  squares  af,  fd,  and  the  two  equal  rect- 
angles ef,  fb  That  is,  the  square  of  ab  is  equal  to  the 
squares  of  ac,  cb,  together  with  twice  the  rectangle  of  ac, 

CB.  ^  Q.    E.    D, 

Corol.    Hence,  if  a  line  be  divided  into  two  equal  parts  ; 
the  square  of  the  whole  line,  will  be  equal  to  four  times  the  ^ 
squa^re  of  half  the  liae. 

THEOREM 


y 

THEOREMS. 


299 


a 

F 

E 

EC 
C 

D 

A.     B 

K.  r 


THEOREM  XXXII. 

The  Square  of  the  Difference  of  two  Lines,  ie  less  than 
the  Sum  of  their  Squares,  by  Twice  the  Rectangle  of  the 
said  Lines. 

Let  AC,  Bc,  be  any  two  li|res,  and  ab 
their  difference  :  then  will  the  square  of 
AB  be  less  than  the  squares  of  ac,  bc,  by 
twice  the  rectangle  of  ac  and  bc.     'Or, 

AB^    =  AC^    -{-  BC2    -»   2 AC  .  BC. 

For  let  ABDE  be  the  square  on  the  dif- 
ference AB,  and  ACFG  the  square  on  the 
line  AC.     Produce  ed  to  h  ;  also  produce 
DB  and  Hc,  and  draw   ki,  making  bi  the  square  of  the  other 
line  BC. 

Now  it  is  visible  that  the  square  ad  is  less  than  the  twcr 
squares  af,  bi,  by  the  two  rectangles  ef,  di.  But  gf  is  equal 
to  the  one  line  ac,  and  ge,  or  fh  is  equal  to  the  other  line  bo; 
consequently  the  rectani^le  ef,  contained  under  eg  and  gf,  is 
equal  to  the  rectangle  of  ac  and  bc. 

Again,  fh  being  equal  to  ci  or  bc  or  dh,  by  adding  the 
common  part  hc,  the  whole  hi  will  He  equal  to  the  whole  fc, 
or  equal  to  ac  ;  and  consequently  the  figure  i>i  is  equal  to  the 
rectangle  contamed  by  ac  and  bc. 

Hence  the  two  figures  ef,  di,  are  two  rectangles  of  the 
two  lines  ac,  bc  ;  and  consequently  the  square  of  ab  is  less 
than  the  squares  of  ac,  bc,  by  twice  the  rectangle  ac  .  bc 

ft.    E.    D. 


THEOHEM  XXXni. 


The  Rectangle  under  the  Sum  and  Diff<p»rence  of  tw© 
Lines,  is  equal  to  the  Difference  of  the  Squares  of  those 
Lines. 

Let  ab,  ac,  be  any  two  unequal  lines  ;  E         K  D 

then  will  the  difference  of  the  squares  of  ~ 

ab,  AC,  be  equal  to  a  rectangle  under  G- 
their  sum  and  difference,     "^fhat  is. 


F 


B 


H 


AB2    —   AC2   =   AB   +   AC   .  AB    —  AC.  Ji.         C 

For,  let  ABDE  be  the  square  of  ab,  and 
ACFG  the  square  of  ac.  Produce  db 
till  BH  be  equal  to  ac  ;  draw  hi  parallel 
to  AB  or  ED.  and  produce  fc  both  ways  to 
I  and  K. 

Then  the  difference  of  the  two  squares    ad,  af,  is  evi- 
dently 


sou 


GEOMETRY. 


dently  the  two  rectangles  ef,  kb.  But  the  rectangles  ef,  bi, 
are  equal,  being  contained  under  equal  lines  ;  for  ek  and  bh 
are  each  equal  to  ac,  and  ge  is  equal  to  ce,  being  each  equal 
to  the  difference  between  ab  and  ac,  or  their  equals  ae  and 
AG.  Therefore  the  two  ef,  kb,  are  equal  to  the  two  kb,  bi, 
or  to  the  whole  kh  ;  and  consequently  kh  is  equal  to  the  dif- 
ference of  the  squares  ad,  af.  But  kh  is  a  rectangle  contE^in- 
ed  by  dh,  or  the  sum  of  ab  and  ac,  and  by  kd,  or  the  differ- 
ence of  AB  and  AC  Therefore  the  difference  of  the  squares 
of  AB,  AC,  is  equal  to  the  rectangle  under  their  sum  and  dif- 
ference. €t.   E.   D. 

THEOREM  XXXIV. 


In  any  Right-angled  Triangle,  the  square  of  the  Hypothe- 
nuse,  is  equal  to  the  Sum  of  the  Squares  of  the  other  two 
Sides. 

Let  ABC  be  a  right-angled  triangle, 
having  the  right  angle  c  ;  then  will  the 
square  of  the  hypothenuse  ab,  be  equal 
to  the  sum  of  the  squares  of  the  other 
two  sides  ac,  cb.     Or  ab^    =  ac^  -f- 

BC^. 

For,  on  ab  describe  the  square  ae, 
anl  on  ac,  cb,  the  squares  ag,  bh  ; 
then  draw  ck  parallel  to  ad  or  be  ;  and 
join  Ai,  BF,  CD,  ce.  D      K    El 

Now,  because  the  line  ac  meets  the  two  cg,  cb,  so  as  to 
make  two  right  angles,  these  two  form  one  straight  line  gb 
(corol.  1,  th.  6).  And  because  the  angle  fac  is  equal  to  the 
angle  dab,  being  ^ch  a  right  angle,  or  the  angle  of  a  square  ; 
to  each  of  these  equals  add  the  common  angle  bac,  so  will 
the  whole  angle  or  sum  fab,  be  equal  to  the  whole  angle  or 
sum  cad.  But  the  line  fa  is  equal  to  the  line  ac,  and  the 
line  ab  to  the  line  ad,  being  sides  of  the  same  square  ;  so 
that  the  two  sides  fa,  ab,  and  their  included  angle  fab,  are 
equal  to  the  two  sidos  ca,  ad,  and  the  contained  angle  cad, 
each  to  each  ;  therefore  the  whole  triangle  afb  is  equal  to  the 
whole  triangle  acd  (th.  1). 

But  the  square  ag  is  double  the  triangle  afb,  on  the  same 
base  fa,  and  between  the  same  parallels  fa,  gb  (th.  26)  ;  in 
like  manner,  the  parallelogram  ak  is  double  the  triangle  acd, 
on  the  same  base  ad,  and  between  the  same  parallels  ad,  ck. 
And  since  the  doubles  of  equal  things,  are  equal  (by  ax.  6)  ; 
therefore  the  square  ag  is  equal  to  the  parallelogram  ak. 


THEOREMS.  301 

In  like  manner,  the  other  square  bh  is  proVed  equal  to  the 
other  parallelogram  bk.  Consequently  the  two  squares  ag 
and  BH  together,  are  equal  to  the  two  parallelograms  ak  and 
BK  together,  or  to  the  whole  square  ae.  That  is,  the  sum  of 
the  two  squares  on  the  two  le§fe  sides,  is  equal  to  the  square 
on  the  greatest  side.  q.  e.  d. 

Corol.  1.  Hence  the  square  of  either  of  the  two  less  sides, 
is  equal  to  the  difference  of  the  squares  of  the  hypothenuse 
and  the  other  side  (ax.  3)  ;  or,  equal  to  the  rectangle  contain- 
ed by  the  sum  and  diflference  of  the  said  hypothenuse  and 
other  side  (th.  33). 

Corol.  2.  Hence  also,  if  two  right-angled  triangles  have 
two  sides  of  the  one  equal  to  two  corresponding  sides  of  the 
other ;  their  third  sides  will  also  be  equal,  and  the  triangles 
identical. 

THEOREM  XXXV. 

In  any  Triangle,  the  Difference  of  the  Squares  of  the 
two  Sides,  is  Equal  to  the  Difference  of  the  Squares  of  the 
Segments  of  the  Base,  or  of  the  two  Lines,  or  Distances, 
included  between  the  Extremes  of  the  Base  and  the  perpen- 
dicular. 

Let  ABC  be  any  triangle,  having 
CD  perpendicular  to  ab  ;  then  will 
the  difference  of  the  squares  of  ac, 
Bc,   be  equal   to  the  difference  of  ^  /  l 

the  squares  of  ad,  bd  ;  that  is,  ac  ^  y      I    a 

— Bc2  =  ad2  -bd2  a    B  3)  a.    D  B 

For,  since  ac^  is  equal  to  ad^  -f-  cd^  >       ,,      ,        . 

and  Bc2  is  equal  to  bd^  -j-  cd^.  J      ^  ^     *      /  ' 
Theref.  the  difference  between  ac^  and  bc^, 
is  equal  to  the  difference  between  ad^  -f-  cd^ 
and  bd2  -f-  cd^, 
or  equal  to  the  difference  between  ad^  and  bd^, 
by  taking  away  the  common  square  cd^  q.  e.  d 

Corol.  The  rectangle  of  the  sum  and  difference  of  the 
two  sides  of  any  triangle,  is  equal  to  the  rectangle  of  the 
sum  and  difference  of  the  distances  between  the  perpendi- 
cular and  the  two  extremes  of  the  base,  or  equal  to  the  rect- 
angle of  the  base  and  the  difference  or  sum  of  the  segments, 
according  as  the  perpendicular  falls  within  or  without  the  tri- 
angle. 

That 


30^ 


GEOMETRY. 


That  is,  AC  -f-  bc  .  ac  —  bc  =  ad  -{-  bd  .  ad  — -  bd 


Or,  AC  -{-  BC  .  AC  —  BC  =  AB  .  AD  —  BD  in  the  2d  figure- 
And  AC  -f-  BC  .  AC  —  BC  =  AB  .  Ao  -|"  BD  in  the  1st  figure. 


THEOREM  XXXVI. 

In  any  Obtuse-angled  Triangle,  the  Square  of  the  Side  sub- 
tending the  Obtuse  Angle,  is  Greater  than  the  Sum  of  the 
Squares  of  the  other  two  Sides,  by  Twice  the  Rectangle  of 
the  Base  and  the  Distance  of  the  Perpendicular  from  the  Ob- 
tuse Angle. 

Let  ABC  be  a  triangle,  obtuse  angled  at  b,  and  cd  perpen  ~ 
dicular  to  ab  ;  then  will  the  square  of  ac  be  greater  than  the 
-  squares  of  ab,  bc,  by  twice  the  rectangle  of  ab,  bd.  That  is, 
Ac2  =  AE^  -\-  Bc2  -{-  2ab  .  BD.  See  the  1st  fig.  above  or  be- 
low. 

For,  since  the  square  of  the  whole  line  ad  is  equal  to  the 
squares  of  the  parts  ab,  bd,  with  twice  the  rectangle  of  the 
same  parts  ab,  wd  (th.  31)  ;  if  to  each  of  these  equals  there 
be  added  the  square  of  cd,  then  the  squares  of  ad,  cd,  will 
be  equal  to  the  squares  of  ab,  bd,  cd,  with  twice  the  rectan- 
gle of  ab.  bd  (by  ax.  2). 

But  the  squares  of  ad,  cd,  are  equal  to  the  square  of  ac  ; 
and  the  squares  of  bd,  cd,  equal  to  the  square  of  bc  (th.  34)  ; 
therefore  the  square  of  ac  is  equal  to  the  squares  of  ab,  bc> 
together  with  twice  the  rectangle  of  ab,  bd.  ^.  e.  jk 

THEOREM  XXXVII. 


In  any  Triangle,  the  Square  of  the  Side  subtending  an 
Acute  Angle,  is  Less  than  the  Squares  of  the  Base  and  the 
other  Side,  by  Twice  the  Rectangle  of  the  Base  and  the  Dis- 
tance of  the  Perpendicular  from  the  Acute  Angle. 

Let  ABC  be  a  triangle,  having 
the  angle  a  acute,  and  cd  perpen- 
dicular to  AB  ;  then  will  the  square 
of  BC,  be  less  than  the  squares  of 
AB,  AC,  by  twice  the  rectangle  of 
AB,  AD.     That  is,  Bc^  =  AB^  -h 

AC2  2aB  .  AD. 


A.    B  D   ,A.     2)  J3 


For, 


THEOREMS. 


SOS 


For,  in  fig.  1,  ac^  is  =  bc^  -f-  ab^  -f-  2ab  .  bd  (th.  36). 
To  eaoii  of  these  equals  add  the  square  of  ab, 
then  is  ab^  -f  ac^  =  bc'^  -f  2ab-^  -f  ^ab  .  ed  (ax.  2), 
or  =  Bc3  4.  2ab  .  AD  (th.  30). 

^.   E.  D. 

Again,  in  fig.  2.  ac^  is  =  ad^  +  dc^  (th.  34). 
And  ab^  =  ads  -{-  db^  -f-  2ad  .  db  (th.  31). 
Theref.AB2  -j-  ac2  =  bd^  -{-  dc^  -f  Sad^  4-  2ad  .  bb  (ax.  2), 
.  or  =  Bc3  4-  2ad  2  4  2ad  .  db  (th.  34), 
or  =  BC2   4  2aB  .  AD  (th.  30).  'q.   E.  J3. 

THEOREM  XXXVIII. 


In  any  Triangle,  the  Double  of  the  Square  of  a  Line  drawn 
from  the  Vertex  to  the  Middle  of  the  Base,  together  with 
Double  the  Square  of  the  half  Base,  is  Equal  to  the  Sum  of  the 
Squares  of  the  other  Two  Sides. 

Let  ABC  be  a  triangle,  and  cd  the  lirie 
drawn  from  the  vertex  to  the  middle  of 
the  base  ab,  dividing  it  into  two  equal 
parts  ad,  db  ;  then  will  the  sum  of  the 
squares  of  ac,  cb,  be  equal  to  twice  the 
sum  of  the  squares  of  c^,  bd  ;  or  ac^  4 
•gb^  =  2cd2  4  2db3. 

For,   let   CE    be    perpendicular   to    the 


A     DEB 

base  ab.      Then, 
since  (by  th   36)  ac^  exceeds  the  sum  of  the  two  squares  ad^ 


and  cd3  (or  bd^  and  cd^)  by  the  double  rectangle  2ad  .  de 
(or  ^bd  ,  de)  ;  and  since  (by  th.  37)  bc^  is  less  than  the  same 
0Mm  by  the  said  double  rectangfe  ;  it  is  manifest  that  both  ac* 
and  Bc^  together  must  be  equal  to  that  sum  twice  taken  ;  the 
excess  on  the  one  part  making  up  the  defect  on  the  other. 

Q.    E.    15. 


THEOREM  XXXIX. 


In  an  Isosceles  Triangle,  the  Square  of  a  Linfe  drawn  fro» 
the  Vertex  to  any  Point  in  the  Base,  together  with  the  Rectan- 
gle of  the  Segments  of  the  Base,  is  equal  to  the  Square  of  one 
of  the  Equal  Sides  of  the  Triang  e. 

Let  ABC  be  the  isosceles  triangle,  and 
CD  a  line  drawn  from  the  vertex  to  any 
point  D  in  the  base  :  then  will  the  square 
of  AC,  be  equal  to  the  square  of  cd,  to- 
gether with  the  rectangle  of  ac  and  db. 

That,  is  Aca  =  cd=  4-  ad  .  db.  

For,     AD:B 


304  GEOMETRY. 

For,  let  cE  bisect  the  vertical  ano;le  ;  then  will  it  also 
bisect  the  base  ab  perpendicularly  making  ae  =  eb  (cor.  1, 
th.  3). 

But,   in  the  triangle  acd,  obtuse  angled  at  d,   the  square 

AC2  is  =  CD2   _|-  AD^-f-  2aD  .  DE  (th.  36), 

or  =  cd2  +  ad  .  AD  4-  2de  (th.  30), 

Ot  =  CD2  -f-  ad  .  AE  4"  DE, 
or  =  CD^  -{-  ad  .  BE  -f  DE, 
or  =  CD^    -f-  ad  .  DB,  A 


ft.  E.  D.    AD£     33 
THEOREM  XL. 

In  any  Parallelogram,  the  two  Diagonals  Bisect  each  other  ; 
and  the  Sum  of  their  Squares  is  equal  to  the  Sum  of  the 
Squares  of  all  the  Four  sides  of  the  Parallelogram. 

Let  ABCD  be  a  parallelogram,  whose 
diagonals  intersect  each  other  in  e  :  then 
will  AE  be  equal  to  ec,  and  be  to  ed  ;  and 
the  sum  of  the  squares  of  ac,  bd, 
will  be  equal  to  the  sum  of  the  squares 
•f  AB,  Bc,  CD,  DA.     That  is, 

AE  =  EC,  and  BE  =  ED, 

and  Ac2  4-  bd2  =  ab^  -|-  bc^  -f  cd^  -}-  da^. 
For,  the  triangles  aeb,  dec,  are  equiangular,  because  they 
have  the  opposite  angles  at  e  equal  (th.  7),  and  the  two  lines 
AC,  BD,  meeting  the  parallels  ab,  dc,  make  the  angle  bae  equal 
to  the  angle  dce,  and  the  angle  abe  equal  to  the  angle  cde, 
and  the  side  ab  equal  to  the  side  dc  (th.  22)  ;  therefore  these 
two  triangles  are  identical,  and  have  their  corresponding  sides 
equal  (th.  2),  viz.  ae  =  ec,  and  be  =  ed. 

Again,  since  ac  is  bisected  in  e,  the  sum  of  the  squares  ad^ 
-f  Dc2  =  2ae2  -f  2de3  (th.  38). 

In  like  manner,  ab^  -j-  bc^  =  2ae2  -f-  2be2  or  2de2  , 

Theref.  ab^  -f  bc2  -j-  cd^  -f  da3  =  4ae2  +  4de2  (ax.  2). 

But,  because  the  square  of  a  whole  line  is  equal  to  4  times 
the  square  of  half  the  line  (cor.  th.  31),  that  is,  ac^  =  4ae2, 
and  bd2  =  4de2. 

Theref.  ab^  -f-  bc^  -{-  cd^  +  t)a^  =  ac^  -f-  bd^  (ax.  1). 

Q.  E,  d. 
THEOREM 


THEOREMS.  306 


THEOREM  XLI. 

ip  a  Line,  drawn  through  or  from  the  Centre  of  a  Circle, 
Bisect  a  Chord,  it  will  be  Perpendicular  to  it  ;  or  if  it  be 
Perpendicular  to  the  Chord,  it  will  Bisect  both  the  Chord 
and  the  arc  of  the  Chord. 

Let  AB  be  any  chord  in  a  circle,  and  cd 
a  line  drawn  from  the  centre  c  to  the 
chord.  Then,  if  the  chord  be  bisected  in 
the  point  d,  cd  will  be  perpendicular  to 

AB. 

For,  draw  the  two  radii  ca,  cb.  Then, 
the  two  triangles  acd,  bcd,  having  ca 
equal  to  cb  (def.  45),  and  cd  comnaon,  also 
AD  equal  to  db  (by  hyp.)  ;  they  have  all  the  three  sides  of 
the  one,  equal  to  all  the  three  sides  of  the  other,  and  so  have 
their  angles  also  equal  (th  5).  Hence  then,  the  angle  adc 
being  eqiial  to  the  angle  bdc,  these  angles  are  right  angles, 
and  the  line  cd  is  perpendicular  to  ab  (def.  II). 

Again,  if  cd  be  perpendicular  to  ab,  then  will  the  chord 
AB  be  bisected  at  the  point  d,  or  have  ad  equal  to  db  ;  and 
the  arc  aeb  bisected  in  the  point  e,  or  have  ae  equal  eb. 

For,  having  drawn  ca,  cb,  as  before.  Then,  in  the  tri- 
angle ABC,  because  the  side  ca  is  equal  to  the  side  cb,  their 
opposite  angles  a  and  b  are  also  equal  (th.  3)  Hence  then, 
in  the  two  triangles  acd,  bcd,  the  angle  a  is  equal  to  the 
angle  b,  and  the  angles  at  d  are  equal  (def  U)  ;  therefore 
their  third  angles  are  also  equal  (corol.  1,  th  17).  And 
having  the  side  cd  common,  they  have  also  the  side  ad  equal 
to  the  side  db  (th.  2). 

Also,  since  the  angle  ace  is  equal  to  the  angle  bce,  the 
arc  AE,  which  measures  the  former  (def.  57),  is  equal  to  the 
arc  BE,  which  measures  the  latter,  since  equal  angles  must 
have  equal  measures. 

Corol.  Hence  a  lirfe  bisecting  any  chord  at  right  angles, 
passes  through  the  centre  of  the  circle. 

THEOREM  XLIL 

If  More  than  Two  Equal  Lines  can  be  drawn  from  *oy 
Point  within  a  Circle  to  the  Circumference,  that  Point  will  be 
the  centre. 

Vor..  k  4e  Ipet 


306 


GEOMKTRY. 


Let  ABC  be  a  circle,  and  d  a  point 
within  it  :  then  if  three  lines,  da,  bb, 
DC,  drawafrom  the  point  d  to  the  cir- 
cumference, be  equal  to  each  other, 
the  point  d  wil!  be  the  centre. 

For,  draw  the  chords  ab.  bc,  which 
let  be  hisected  in  the  point  e,  f,  and 

join  DE,  DF. 

Then,  the  two  triangles,  dae,  dbe,  have  the  side  da  equal 
to  the  side  db  by  supposition,  and  the  side  ae  equal  to  the 
side  eb  by  hypothesis,  also  the  side  de  common  :  therefore 
these  two  triangles  are  identical,  and  have  the  angles  at  b 
equal  to  each  other  (th.  6)  ;  consequently  de  is  perpendicu- 
lar to  the  n»iddle  of  the  chord  ab  (def  11),  and  therefore 
passes  through  the  centre  of  the  circle  (corol.  th.  41). 

In  like  manner,  it  may  be  shown  that  df  passes  through 
the  centre.  Consequently  the  point  d  is  the  centre  of  the 
circle,  and  the  three  equal  lines,  da,  db,  dc,  are  radii. 

q.   E.    D. 


THEOREM  XUir. 


If  two  Circles  touch  one  another  Internally,  the  Centres  of 
the  Circles,  and  the  Point  of  Contact  will  be  all  in  the  Same 
Right  Line. 

L*»t  the  two  circles  abc,  ade,  touch 
one  another  internally  in  the  point  a  ; 
then  will  the  point  a  and  the  centres  of 
those  circles  be  all  in  the  same  right 
line. 

For,  let  F  be  the  centre  of  the  circle 
ABC,  through  which'  draw  the  diameter 
AFC.  Then,  if  the  centre  of  the  other 
circle  can  be  out  of  this  line  ac,  let  it  bp 
supposed  in  some  other  point  as  g  ;  through  which  draw  the 
line  FG  cutting  the  two  circles  in  b  and  d. 

Now,  in  tjhe  triangle  afg,  the  sum  of  the  two  sides,  fg, 
GA,  is  greater  than  the  third  side  af  (th  JO),  or  greater  than 
its  equal  radius  fb.  From  each  of  these  take  away  the 
common  part  fg,  and  the  remainder  ga  will  be  greater 
than  the  remainder  gb.  But  the  point  g  being  supposed 
the  centre  of  the  inner  circle,  its  two  radii,  ga,  gd,  are 
equal  to  each  other  ;  consequently  gd  will  also  be  greater 
than  gb.     But  ade  being  the  inner  circle,  gd  is  necessarily 

less 


THEOREMS. 


307 


less  than  gb.  So  that  od  is  both  greater  and  less  than  gb  ; 
which  is  absurd.  Consequently  the  centre  g  cannot  be  out  of 
the  line  ajc.  q.  e.  d. 

I'HEOREM  XLIV. 


If  two  Circles  Touch  one  another  Externally,  the  Centres 
of  the  Circles  and  the  Point  of  Contact  will  be  all  in  the  Same 
Right  Line. 

Let  the  two  circles  abc,  ade,  touch  one 
another  externally  at  the  point  a  ;  then  will 
the  point  of  contact  a  and  the  centres  of  the 
two  circlejj  be  all  in  the  same  right  line. 

For,  let  F  be  the  centre  of  the  circle  abc, 
through  which  draw  the  diameter  afc,  and 
produce  it  to  the  other  circle  at  e  Then,  if 
the  centre  of  the  other  circle  ade  can  be  out 
of  the  lineFE,  letit,  if  possible,  be  supposed 
in  some  other  point  as  g  ;  and  draw  the  lines 
AG,  FB,  DG,  cutting  the  two  circles  in  b  and  d. 

Then,  in  the  triani^le  afg,  the  sum  of  the  two  sides  af, 
AG,  is  greater  than  the  third  .«ide  fg  (th.  10)  But,  f  and  g 
being  the  centres  of  the  two  circles,  the  two  radii  ga,  gd, 
are  equal,  as  are  also  the  two  radii  af,  fb.  Hence  the  sum 
of  ga,  af,  is  equal  to  the  sura  of  gd,  bp  ;  and  therefore  this 
latter  sum  also,  gd,  bf,  i.^  greater  than  gf,  which  is  absurd. 
Consequently  the  centre  o  cannot  be  out  of  the  line  ef. 

Q.    E.    D, 


THEOREM  XLV. 


Any  Chords  in  a  Circle,  which  are  Equally  Distant  from  the 
Centre,  are  Equal  to  each  other  ;  or  if  they  be  Equal  to  each 
other,  they  will  be  Equally  Distant  from  the  Centre. 

Let  AB,  CD,  be  any  two  chords  at  equal 
distances  from  the  centre  g  :  then  will 
these  two  chords  ab,  cd,  be  equaltoeach 
other. 

For,  draw  the  two  radii  ga,  gc,  and 
the  two  perpendiculars  ge.  gf,  which  are 
the  equal  distances  from  the  centre  g, 
Then,  the  two  right  angled  trianj^les. 
gae,  gcf,  having  the  side  ga  equal  the  side  gc,  and  the  side' 
GE  equal  the  side   gf,  and   the  angle  at  e  equal  to  the  an^-Ie 

''at 


308 


GEOMETRY. 


at  F,  therefore  the  two  triangles  gae, 
GCF,  are  identical  (cor.  2,  th.  34),  and 
have  the  hne  ae,  equal  the  line  cf. 
But  AB  is  the  double  of  ae,  and  cd  is 
the  double  of  of  (th.  41)  ;  therefore 
^B  is  equal  to  cd  (by  ax.  6).      ft.  e.  d. 

Again,  if  the  chord  ab  be  equal  to 
the  chord  cd  :  then   will  their  distances  from  the  centre,  ge^ 
GF,  also  be  equal  to  each  other. 

For,  since  ab  is  equal  cd  by  supposition,  the  half  ae  is 
equal  the  half  cf.  Also  the  radii  ga,  gc,  being  equal,  as  well 
as  the  right  angles  e  and  f,  therefore  the  third  sides  are  equal 
(cor.  JjJ,  th.  34),  or  the  distance  ge  equal  the  distance  gf. 

ft.  e.  d. 


THEOREM  XLVl. 


A  Line  Perpendicular  to  the  Extremity  of  a  Radius,  is  a  Tan- 
gent to  the  Circle. 

Let  the  line  adb  be  perpendicular  to  the 
radius  cd  of  a  circle  ;  then  shall  ab  touch 
the  circle  in  the  point  D  only. 

For,  from  any  other  point  e  in  the  line 
ab,  draw  cfe  to  the  centre,  cutting  the 
circle  in  f. 

Then,  because  the  angle  i),  of  the  trian- 
gle CDE,  is  a  right  angle,  the  angle  at  e  is  acute  (th.  17,  cor.  3), 
and  consequently  less  than  the  angle  d.  But  the  greater 
side  is  always  opposite  to  the  greater  angle  (th.  9)  ;  therefore 
the  side  ce  is  greater  than  the  side  cd,  or  greater  than  its 
equal  cf.  Hence  the  point  e  is  without  the  circle  :  and  the 
same  for  every  other  point  in  the  line  ab.  Consequently  the 
whole  line  is  without  the  circle,  and  meets  it  in  the  point  9 
only. 


THEORE 


THEOREMS.  309 


THEOREM  XLVn. 

When  a  Line  is  a  Tangent  to  a  Circle,  a  Radiufs  drawn  t© 
the  Point  of  Contact  is  Perpendicular  to  the  Tangent. 

Let  the  line  ab  touch  the  circumference  of  a  circle  at  the 
point  D  ;  then  will  the  radius  cd  be  perpendicular  to  the 
tangent  ab.     [See  the  last  figure.] 

For,  the  line  ab  being  wholly  without  the  circumference 
except  at  the  point  d,  every  other  line,  as  ce  drawn  from 
the  centre  c  to  the  line  ab,  mu«t  pass  out  of  the  circle  to 
arrive  at  this  line.  The  line  cd  is  therefore  the  shortest  that 
can  be  drawn  from  the  point  c  to  the  line  ab,  and  consequent- 
ly (th.  21)  it  is  perpendicular  to  that  line. 

Corol.  Hence,  conversely,  a  line  ^rawn  perpendicular  to  a 
tangent,  at  the  point  of  contact,  passes  through  the  centre  o£ 
the  circle. 

THEOREM  XLVin. 

The  Angle  formed  by  a  Tangent  and  Chord  is  Measured  by 
Half  the  Arc  of  that  Chord. 

Let  AB  be  a  tangent  to  a  circle,  and  cd 
a  chord  drawn  from  the  point  of  contact  c  ; 
then  is  the  angle  bcd  measured  by  half  the 
arc  CFD,  and  the  angle  acd  measured  by 
half  the  arc  cgd. 

For,  draw  the  radius  ec  to  the  point  of 
contact,  and  the  radius  ef  perpendicular  to 
the  chord  at  h. 

Then  the  radius  ef,  being  perpendicular  to  the  chord  cd, 
bisects  the  arc  cfd  (th.  4  J).     Therefore  cf  is  half  the  arc 

CFB. 

In  the  triangle  ceh,  the  angle  h  being  a  right  one,  the  sum 
of  the  two  remaining  angles  e  and  c  is  equal  to  a  right  angle 
(corol.  3,  th.  17),  which  is  equal  to  thfe  angle  bce,  becau!>e 
the  radius  ce  is  perpendicular  to  the  tangent  From  each  of 
these  equals  take  away  the  common  part  or  angle  g,  and  there 
remains  the  angle  e  equal  to  the  angle  bcd.  But  the  angle 
E  is  measured  by  the  arc  cf  (def  67),  which  is  the  half  of  cfd; 
therefore  the  equal  angle  bcd  must  also  have  the  same  mea- 
sure, namely,  half  the  arc  cfd  of  the  chord  cd. 

Again, 


310  GEOMETRY. 

Again,  tbe  line  oef,  being  perpendicular 
to  the  chord  cd,  bisects  the  arc  cgd,  (th. 
41).  Therefore  cg  is  half  the  arc  cgd. 
Now,  since  the  line  ce,  meeting  fg,  makes 
the  sum  of  the  two  angles  at  e  equal  to 
two  right  angles  (th.  6),  and  the  line  cd 
makes  with  ab  the  sum  of  the  two  angles 
at  c  equal  to  two  right  angles  ;  if  from 
these  two  equal  sums  there  be  taken  away  the  parts  or  angles 
CEH  and  BCH  which  have  been  proved  equal,  there  remains 
the  angle  ceg  equal  to  the  angle  ach.  But  the  former  of 
these,  ceg,  being  an  angle  at  the  centre,  is  measured  by  the 
arc  CG  (def.  67)  ;  consequently  the  equal  angle  acd  must 
also  have  the  same  measure  cg,  which  is  half  the  arc  cgd  of 
the  chord  cd.  q.  ,e.  d. 

Corol.  1.  The  sum  of  two  right  angles  is  measured  by 
half  the  circumference.  For  the  two  angles. bcd,  acd,  which 
make  up  two  rijsht  angles,  are  measured  by  the  arcs,  cf,  cg, 
which  make  up  half  the  circumference,  fg  being  a  diameter. 

Corol.  2.  Hence  also  -one  right  angle  must  have  for  its 
measure  a  quarter  of  the  circumference,  or  90  degrees. 


THEOREM  XUX. 


An  Angle  at  the  Circumference  of  a  Circle,  is  measured  by 
Half  the  Arc  that  subtends  it. 

Let  BAC  be  an  angle  at  the  circumference  ; 
it  has  for  its  measure,  hall  the  arc  bc  which 
subtends  it. 

For,  suppose  the  tangent  de  passing 
through  the  point  of  contact  a.  Then,  the 
angle  dac  being  measured  by  half  the  arc 
ABC,  and  the  angle  dab  by  half  the  arc  ab 
(th.  48y  ;  it  follows,  by  equal  subtraction  that  the  difference, 
or  angle  BAC,  must  be  measured  by  half  the  arc  bc,  which  it 
stands  upon.  Q.  £.  d. 


THEOREM 


THEOREMS. 


311 


THEOREM  L. 

All  Angles  in  the  Same  Segment  of  a  Circle,  or  Standing  on 
the  Same  Arc,  are  equal  to  each  other. 

Let  0  and  d  be  two  angles  in  the  same 
segment  acdb,  or,  which  is  the  same  thing, 
St mdingon  the  supplemental  arc  aeb  ;  then 
will  the  angle  c  be  equal  to  the  angle  d. 

For  each  of  these  angles  is  measured  by 
half  the  arc  aeb  ;  and  thus,  having  equal 
measures,  they  are  equal  to  each  other  (ax. 
U). 

THEOREM  LI. 

An  Angle  at  the  Centre  of  a  Circle  is  Double  the  Angle  at  the 
Circumference,  when  both  stand  on  the  Same  Arc. 

Let  c  be  an  angle  at  the  centre  c,  and 
D  an  angle  at  the  circumference,  both  stand- 
ing on  the  same  arc  or  same  chord  ab  :  then 
will  the  angle  c  be  double  of  the  angle  d, 
or  the  angle  d  equal  to  half  the  angle  g. 

For,  the  angle  at  the  centre  c  is  measur- 
ed by  the  whole  arc  aeb  (def.  57),  and  the 
angle  at  the  circumference  d  is  measured  by  half  the  same 
arc  aeb  (th.  49)  ;  therefore  the  angle  d  is  only  half  the  angle 
c,  or  the  angle  c  double  the  angle  d. 


THEOREM  UI. 


An  Angle  in  a  Semicircle,  is  a  Right  Angle. 


If  ABC  or  ADC  be  a  semicircle  ;  then 
any  angle  d  in  that  semicircle,  is  a  right 
angle. 

For,  the  angle  d,  at  the  circumference, 
is  measured  by  half  the  arc  abc  (th.  49), 
that  is,  by  a  quadrant  of  the  circumference. 
But  a  quadrant  is  the  measure  of  a  right 
an^le  (corol.  4.  th.  6  ;  or  corol.  2,  th.  48). 
angle  d  is  a  right  angle. 


O 

B 
Therefore  the 


THEOREM 


312  GEOMETRY. 


THEOUfiM  UII.^ 

The  Angle  formed  by  a  Tangent  to  a  Circle,  and  a  Cheri 
drawn  from  the  Point  of  Contact,  is  Equal  to  the  Angle  in  the 
Alternate  Segment. 

If  AB  be  a  tangent,  and  ac  a  chord,  and  , 

D  any  anj^e  in  the  alternate  segment  adc  ; 
then  will  the  angle  d  be  equal  to  the  angle 
BAG  made  by  the  tangent  and  thord,  of  the 
arc  AEc. 

For  the  angle  d,  at  the  circumference, 
is  measured  by  lialf  the  arc  aec  (th.  49)  ; 
and  the  angle  bac,  made  by  the  tangent  and  chord,  is  als» 
measured  by  the  same  half  ar«  aec  (th.  48)  ;  therefore  these 
two  angles  are  equal  (ax.  11). 

THEOREM  LIV. 

The  Sum  of  any  Two  Opposite  Angles  of  a  Quadrangle  In- 
scribed in  a  Circle,  is  Equal  to  Two  Right  Angles. 

Let  ABCD  be  any  quadrilateral  inscribed 
in  a  circle  ;  then  shall  the  sum  of  the  two 
opposite  angles  a  aud  c,  or  b  and  d,  be 
equal  to  two  right  angles. 

For  the  angle  a  is  measured  by  half  the 
arc  ©CB,  which  it  stands  on,  and  the  angle' 
c  by  half  the  arc  dab  (th.  49)  ;  therefore 
the  sum  of  the  two  angles  a  and  c  is  measured  by  half  the 
sum  of  these  two  arcs,  that  is,  by  half  the  circumference. 
But  half  the  circumference  is  the  measure  of  two  right  an- 
gles (corol.  4,  th.  6)  ;  therefore  the  sum  of  the  two  opposite 
angles  a  and  c  is  equal  to  two  right  angles.  In  hk^  manner 
it  is  shown,  that  the  sum  of  the  other  two  opposite  angles,  b 
and  B,  is  equal  tp  two  right  angles.  Q.  e.  d. 

THEOREM  LV. 

If  any  Side  of  a  Quadrangle,  Inscribed  in  a  Circle,  be  Pro- 
duced out,  the  Outward  Angle  will  be  Equal  to  the  InwardI 
Opposite  Angle. 

If  the  side  ab,  of  the  quadrilateral  abcd, 
inscribed  in  a  circle,  be  produced  to  e  ;  the 
outward  angle  DAE.will  be  equal  to  the 
inward  opposite  angle  c. 

For, 


THEOREMS.  '  313 

For,  the  sum  of  the  two  adjacent  angles  dae  and  dab  is 
equal  to  two  right  angles  (th.  6)  ;  and  the  sum  of  the  two 
opposite  angles  c  and  dab  is  also  equal  to  two  right  angles 
(th.  S4)  ;  therefore  the  former  sum,  of  the  two  angles  dae 
and  dab,  is  equal  to  the  latter  sum,  of  the  two  c  and  da» 
(ax.  1).  From  each  of  these  equals  taking  away  the  common 
angle  dab,  there  remains  the  angle  dae  equal  the  angle  c. 

^.  B.  D. 


THEOREM  LVL 


Any  Two  Parallel  Chords  Intercept  Equal  Arcs. 

Let  the  two  chords  ab,  cd,  be  parallel  : 
then  will  the  arcs  ac,   bd,    be   equal  j  or 

AC   =  BD. 

For»  draw  the  line  bc.  Then,  because 
the  Hues  ab,  gd,  are  parallel,  the  alternate 
angles  b  and  c  are  equal  (th.  12).  But  the 
angle  at  the  circumference  b,  is  measured  by  half  the  arc 
AC  (th.  49)  ;  and  the  other  equal  angle  at  the  circumference 
c  is  measured  by  half  the  aire  bd  :  therefore  the  halves  of  the 
arcs  AC,  BD,  and  consequently  the  arcs  themselves,  are  also 
„  equal.  <l.  E.  D. 

THEOREM  LVn. 

When  a  Tangent  and  Chord  are  Parallel  to  each  other,  they 
Intercept  Equal  Arcs. 

Let  the  tangent  Ape  be  parallel  to  the 
'  jchord  DF  ;  then  are  the  arcs  bd,  bf,  equal ; 
that  is,  BD  =  BF. 

For,  draw  the  chord  bd.  Then,  be- 
cause the  Unes  ab,  df,  are  parallel,  the  al- 
ternate angles  d  and  b  are  equal  (th.  12). 
But  the  angle  b,  formed  by  a  tangent  and  chord,  is  measured 
by  half  the  arc  bd  (th.  48)  ;  and  the  other  angle  at  the  circum- 
ference D  is  measured  by  half  the  arc  bf  (th.  49)  ;  there- 
fore the  arcs  bd,  bf,  are  equal.  q.  e.  d. 


Vol.  I.  41  THEQRBM 


314  GEOMETRY. 


THEOREM  LVra. 


The  Angle  formed,  Within  a  Circle,  by  the  Intersection  of  two 
Chords,  is  Measured  by  Half  the  Sum  of  the  Two  Inter- 
cepted Arcs. 

Let  the  two  chords  ab,  cd,  intersect  at 
the  point  e  :  then  the  angle  aec,  or  deb,  is 
measured  by  half  the  sum  of  two  arcs  ac, 

DB. 

For,   draw  the  chord  af  parallel  to  cd. 
Then,  because  the  lines  af,  cd,  are  parallel, 
and  ab  cuts  them,  the  angles  on  the  same 
side  A  and  deb  are  equal  (th.  14).     But  the  angle  at  the  cir- 
cumference a  is  measured  by  half  the  arc  bf,  or  of  the  sum  of 
FD  and  DB  (th.  49)  ;  therefore  the  angle  e  is  also  measured  by 
half  the  sum  of  fd  and  db. 

Again,  because  the  chords  af,  cd,  are  parallel,  the  arcs  ac 
fd,  are  equal  (th.  56)  ;  therefore  the  sum  of  the  two  arcs  ac, 
DB  is  equal  tq  the  sum  of  the  two  f  D,  db  ;  and  consequently 
the  angle  e,  which  is  measured  by  half  the  latter  sum,  is  also 
measured  by  half  the  former.  r.  e.  d. 

THEOREM  LIX. 

The  Angle  formed.  Without  a  Circle,  by  two  Secants,  is 
Measured    by  Half  the    Diflference    of  the    Intercepted 

Arcs. 

Let  the  angle  e  be  formed  by  two  se- 
cants eab  and  ecd  ;  this  angle  is  measured 
by  half  the  difference  of  the  two  arcs 
AC,  DB,  intercepted  by  the  two  secants. 

Draw  the  chord  af  parallel  to  cd*  Then, 
because  the  lines  af,  cd,  are  parallel,  and 
AB  cuts  them,  the  angles  on  the  same  side  a 
and  BED  are  equal  (th.  14).  But  the  angle  a,  at  the  circum- 
ference, is  measured  by  half  the  arc  bf  (th.  49),  or  of  the 
difference  of  df  and  db  :  therefore  the  equal  angle  e  is  als» 
measured  by  half  the  difference  of  df,  db. 

Again,  because  the  chords  af,  cd,  are  parallel,  the  arcs 
AC,  FD,  are  equal  (th.  56) ;  therefore  the  difference  of  the 

two 


THEOREMS.  316 

two  arcs  ac,  db,  is  equal  to  the  difference  of  the  two  df,  db. 
Consequently  the  angle  e,  which  is  measured  hy  half  the  latter 
difference,  is  also  measured  by  half  the  former.  ^,  e.  d« 


THEOREM  LX. 

The  Angle  formed  by  Two  Tangents,  is  Measured  by  Half 
the  Difference  of  the  two  Intercepted  Arcs. 

Let  eb,  ed,  be  two  tangents  to  a  circle 
at  the  points  a,  c  ;  then  the  angle  e  is  mea- 
sured by  half  the  difference  of  the  two  arcs, 

^FA,  CGA. 

For,   draw  the  chord  af  parallel  to  ed. 
Then,  because  the  lines  af,  ed,  are  parallel, 
and  EB  meets  them,  the  angles  on  the  same 
side  A  and  e  are  equal  (th.   14).     But  the 
angle  a,  formed  by  the  chord  af  and  the  tangent  ab,  is  mea- 
sured by  half  the  arc  af  (th.  48)  ;  therefore  the  equal  angle 
E  is  als©  measured  by  half  the  same  arc  af,  or  half  the  diffe- 
rence of  the  arcs  cfa  and  cf,  or  cga  (th.  67). 

Corol.  In  like  manner  it  is  proved,  that 
the  angle  e  formed  by  a  tangent  ecd,  and 
a  secant  eab,  is  measured  by  half  the  dif- 
ference of  the  two  intercepted  arcs  ca  and 
cfb. 


l-HEOREM  LXL 

When  two  Lines,  meeting  a  Circle  each  in  two  Points,  Cut  one 
another,  either  Within  it  or  Without  it  ;  the  Rectangle  of 
the  Parts  of  the  one,  is  Equal  to  the  Rectangle  of  the 
Parts  of  the  other  ;  the  Parts  of  each  being  measured  from 
the  point  of  meeting  to  the  two  intersections  with  the  cir- 
cumference. 


l«ET 


316 


GEOMETRY. 


Let  the  two  lines,  ab,  cd,  meet  each 
other  in  e  ;  then  the  rectangle  of  ae,  eb, 
will  be  equal  to  the  rectangle  of  ce,  ed.  Or, 

AE  .   EB    =  CE   .  ED. 

For,  through  the  point  e  draw  the  diame- 
ter FG  ;  also,  from  the  centre  h  draw  the 
radius   dh,  and  draw  hi   perpendicular  to 

CD. 

Then,  since  deh  is  a  triangle,  and  the 
perp.  HI  bisects  the  chord  cd  (th.  41),  the 
line  ce  is  equal  to  the  difference  of  the 
segments  di,  ei,  the  sum  of  them  being 
DE.  Also,  because  h  is  the  centre  of  the 
circle  and  the  radii  dh,  fh,  gh,  are  all  equal,  the  line  eg  is 
equal  to  the  sum  of  the  sides  dh,  he  ;  and  ef  is  equal  to  their 
diflference. 

But  the  rectangle  of  the  sum  and  difference  of  the  two  sides 
of  a  triangle,  is  equal  to  the  rectangle  of  the  sum  and  dif- 
ference of  the  segments  of  the  base  (th.  36)  ;  therefore  the 
rectangle  of  fe,  eg,  is  equal  to  the  rectangle  of  ce,  ed.  In 
like  manner  it  is  proved,  that  the  same  rectangle  of  fe,  eg, 
is  equal  to  the  rectangle  of  ae,  eb.  Consequently  the  rectan- 
gle of  AE,  EB,  is  also  equal  to  the  rectangle  of  ce,  ed  (ax.  1). 

■'■■'''■  Q.    E.    D. 

Corol.  1.  When  one  of  the  lines  in  the 
second  case,  as  de,  by  revolving  about  the 
point  E,  comes  into  the  position  of  the  tan- 
gent EC  or  ED,  the  two  points  c  and  d  run- 
ning into  one  ;  then  the  rectangle  of  ce,  ed, 
becomes  the  square  of  ce,  because  ce  and 
DE  are  then  equal.  Consequently  the  rect- 
angle of  the  parts  of  the  secant,  ae,  eb,  is 
equal  to  the  square  of  the  tangent  ce^  . 

Corol.  2.  Hence  both  the  tangents  ec,  ef,  drawn  from  the 
same  point  e,  are  equal;  since  the  square  of  each  is  equal  to 
the  same  rectangle  or  quantity  ae  .  EB. 


THEOREM  LXn. 

In  Equiangular  Triangles,  the  Rectangles  of  the  Correspond- 
'  ing  or  Like  Sides,  taken  alternately,  are  equal. 


Let 


THEOREMS. 


3Vi 


Let  ABC,  DEP,  be  two  equiangular 
triangles,  having  the  angle  a  =  the 
angle  d,  the  angle  b  =  the  angle  e, 
and  the  angle  c  =  the  angle  f  ;  also 
the  like  sides  ab,  de,  and  ac,  df,  being 
those  opposite  the  equal  angles  :  then 
will  the  rectangle  of  ab,  df,  be  equal 
to  the  rectangle  of  ac,  de. 

In  BA  produced  take  ag  equal  to  dp  ;  and  through  the 
three  points  b,  c,  g,  conceive  a  circle  ^bcch  to  be  described, 
meeting  ca  produced  at  h,  and  join  gh. 

Then  the  angle  g  is-equal  to  the  angle  c  on  the  same  arc 
BH,  and  the  angle  h  equal  to  the  angle  b  on  the  same  arc  cg 
(th.  60)  ;  also  the  opposite  angles  at  a  are  equal  (th.  7)  : 
therefore  the  triangle  agh  is  equiangular  to  the  triangle 
acb,  and  consequently  to  the  triangle  dfe  also.  But  the 
two  like  sides  AG,  df,  are  also  equal  by  supposition;  conse- 
quently the  two  triangles  agh,  dfe,  are  identical  (th.  2), 
having  the  two  sides  ag,  ah,  equal  to  the  two  df,  de,  each  to 
each. 

But  the  rectangle  ga  .  ab  is  equal  to  the  rectangle  ha  .  ac 
(th.  61)  :  consequently  the  rectangle  df  .  ab  is  equal  the 
rectangle  de  .  ac  q.  e.  d. 


THEOREM  LXUL 


The  Rectangle  of  the  two*  Sides  of  any  Triangle,  is  Equal  to 
the  Rectangle  of  the  Perpendicular  on  the  third  Side  and 
the  Diameter  of  the  Circumscribing  Circle. 


Let  CD  be  the  perpendicular,  and  ce 
the  diameter  of  the  circle  about  the  tri- 
angle ABC  ;  then  the  rectangle  ca  .  cb  is 
=  the  rectangle  cd  .  ce. 


For,  join  be  :  then  in  the  two  trian- 
gles acd,  ecb,  the  angles  a  and  e  are 

equal,  standing  on  the  same  arc  bc  (th.  60)  :  also  the  right  an- 
gle D  is  equal  to  the  angle  b,  which  is  also  a  right  angle,  being 
in  a  semicircle  (th.  52)  :  therefore  these  two  triangles  have  also 
their  third  angles  equal,  and  are  equiangular.  Hence,  ac,  ce, 
and  CD,  cb,  being  like  sides,  subtending  the  equal  angles,  the 
rectangle  ac  .  cb,  of  the  first  and  last  of  them,  is  equal  to  the 
rectangle  ce  .cd,  of  the  other  two  (th.  62). 


THEOREM 


318 


GEOMETRY. 


THEOREM  LXIV. 


The  Square  of  a  line  bisecting  any  Angle  of  a  Triangle, 
together  with  the  Rectangle  of  the  Two  Segments  of  the 
opposite  Side,  is  Equal  to  the  Rectangle  of  the  two  other 
Sides  including  the  Bisected  Angle. ♦ 

Let  CD  bisect  the  angle  the  c  of  triangle 
ABC  ;  then  the  square  cd^  -f-  the  rectangle 
AD  .  DB  is  =  the  rectangle  ac  .  cb. 

For,  let  CD  be  produced  to  meet  the  cir- 
cumscribing circle  at  e,  and  join  ae. 

Then  the  two  triangles  ace,  bcd,  are 
equiangular :  for  the  angles  at  c  are  equal 
by  supposition,  and  the  angles  b  and  e  are  equal,  standing  on 
the  same  arc  ac  (th.  50)  ;  consequently  the  third  angles 
at  A  and©  are  equal  (corol.  1,  th  17)  :  also  ac,  cd,  and 
CE,  CB,  are  like  or  corresponding  sides,  being  opposite  to 
equal  angles  ;  therefore  the  rectangle  ac  .  cb  is  =  the 
rectangle  cd  .  ce  (th  62).  But  the  latter  rectangle  cd  .  ce 
is  =  cd2  -f  the  rectangle  cd  .  de  (th  30)  ;  therefore  also 
the  former  rectangle  ac  ,  cb  is  also  =  cd^  -f-  cd  .  de,  or  equal 
to  cd2  -j-  ad  .  db,  since  cd  .  de  is  =  ad  .  db  (th.  61). . 

^.  E.  D. 


THEOREM  LXV. 

The  Rectangle  of  the  two  Diagonals  of  any  Quadrangle  In- 
scribed in  a  Circle,  is  equal  to  the  sum  of  the  two  Rect- 
angles of  the  Opposite  Sides. 

Let  abcd  be  any  quadrilateral  inscribed 
in  a  circle,  and  ac,  bd,  its  two  diagonals  : 
then  the  rectangle  ac  .  bd  is  =  the  rectan- 
gle ab  .  DC  -}-  the  rectangle  AD  .  bc. 


For,  let  ce  be  drawn,  making  the  angle 
BCE  equal  to  the  angle  dca.  Then  the  two 
triangles  acd,  bce,  are  equiangular  ;  for  the  angles  a  and 
B  are  equal,  standing  on  the  same  arc  dc  ;  and  the  angles 
dca,  bce,  are  equal  by  supposition  ;  consequently  the  third 
angles  adc,  bec  are  also  equal ;  also,  ac,  bc,  and  ad,  be,  are 
like  or  corresponding  sides  ;  being  opposite  to  the  equal  an- 
gles :  therefore  the  rectangle  ac  .  be  is  =  the  rectangle  ad  .  bc 
(th.  62V 

Again, 


THEOREMS.  319 

Again,  the  two  triangles  abc,  dec,  are  equiangular  :  for 
the  angles  bag,  bdc,  are  equal,  standing  on  the  same  arc  bc  ; 
and  the  angle  dce  is  equal  to  the  angle  bca,  hy  adding  the 
common  angle  ace  to  the  two  equal  angles  dca,  bce  ;  therefore 
the  third  angles  e  and  abc  are  also  equal  :  but  ac,  dc,  and  ab, 
DE,  are  the  like  sides  :  therefore  the  rectangle  ac  .  de  is  «« 
the  rectangle  ab  .  dc  (th.  62). 

Hence,  by  equal  additions,  the  sum  of  the  rectangles  ac  . 
BE  +  AC  .  DE  is  =  AD  ,  BC  +  ab  .  DC  But  the  formcr  sum 
of  the  rectangles  ac  .  be  +  ac  .  de  is  =  the  rectangle  ac  . 
BD  (th.  30)  :  therefore  the  same  rectangle  ac  .  bd  is  equal  to 
the  latter  sum,  the  rect.   ad  .  bc  4-  the  rect.  ab  .  dc  (ax.  1). 

Q.  E.  D. 


OF  RATIOS  AND  PROPORTIONS. 


DEFINITIONS. 

Def.  76.  Ratio  is  the  proportion  or  relation  which  one 
magnitude  bears  to  another  magnitude  of  the  same  kind  with 
respect  to  quantity. 

Note.  The  measure,  or  quantity,  of  a  ratio,  is  conceived, 
by  considering  what  part  or  parts  the  leading  quantity,  called 
the  Antecedent,  is  of  the  other,  called  the  Consequent  ;  or 
what  part  or  parts  the  number  expressing  the  quantity  of  the 
former,  is  of  the  number  denoting  in  like  manner  the  latter. 
So,  the  ratio  of  a  quantity  expressed  by  the  number  2,  to  a 
like  quantity  expressed  by  the  number  6,  is  denoted  by  6 
divided  by  2,  or  f  or  3  :  the  number  2  being  3  times  con- 
tained in  6,  or  the  third  part  of  it.  In  like  manner,  the  ratio 
of  the  quantity  3  to  6,  is  measured  by  f  or  2  ;  the  ratio  of  4 
to  6  is  I  or  \\  ;  that  of  6  to  4  is  f  or  |  ;  &c. 

77.  Proportion  is  an  equality  of  ratios.     Thus, 

78.  Three  quantities  are  said  to  be  Proportional,  when  the 
ratio  of  the  first  to  the  second  is  equal  to  the  ratio  of  the  second 
to  the  third.  As  of  the  three  quantities  a  (2),  b  (4),  c  (8), 
where  |.  =  |  ==  2,  both  the  same  ratio. 

79.  Four  quantities  are  said  to  be  Proportional,  when  the 
ratio  of  the  first  to  the  second,  is  the  same  as  the  ratio  of  the 
third  to  the  fourth.  As  of  the  four,  a  (2),  b  (4),  c  (6),  d  (10), 
where  4  — .  u  —  2,  both  the  same  ratio. 


320  6E0METRY. 

Note,  To  denote  that  four  quantities,  a,  b,  c,  d,  are  pro- 
portional, they  are  usually  stated  or  placed  thus,  a  :  b  : :  c  :  d  ; 
and  read  thus,  a  is  to  b  as  c  is  to  d.  But  when  three  quanti- 
ties are  proportional,  the  middle  one  is  repeated,  and  they  are 
written  thus,  a  :  b  : :  b  :  c. 

80.  Of  three  proportional  quantities,  the  middle  one  is  said 
to  be  a  Mean  Proportional  between  the  other  two  ;  and  the 
last,  a  Third  Proportional  to  the  first  and  second. 

81.  Of  four  proportional  quantities,  the  last  is  said  to  be  a 
Fourth  Proportional  to  the  other  three,  taken  in  order. 

82.  Quantities  are  said  to  be  Continually  Proportional,  or 
in  Continued  Proportion,  when  the  ratio  is  the  same  between 
every  two  adjacent  terms,  viz.  when  the  first  is  to  the  second, 
as  the  second  to  the  third,  as  the  third  to  the  fourth,  as  the 
fourth  to  the  fifth,  and  so  on,  all  in  the  same  common  ratio. 

As  in  the  quantities  1,  2,  4,  8,  16,  &c. ;  where  the  common 
ratio  is  equal  to  2.  ... 

-  83.  Of  any  number  of  quantities,  a,  b,  c,  d,  the  ratio  of  the 
first.  A,  to  the  last  d,  is  said  to  be  Compounded  of  the  ratios 
of  the  first  to  the  second,  of  the  second  to  the  third,  and  so  on 
to  the  last. 

84.  Inverse  ratio  is,  when  the  antecedent  is  made  the  con- 
sequent, and  the  consequent  the  antecedent. — Thus,  if  1  :  2  : : 
3:6;  then  inversely,  2  :  1  :  :  6  :  3. 

85.  Alternate  proportion  is,  when  antecedent  is  compared 
with  antecedent,  and  consequent  with  consequent, — As,  if 
1  :  2  :  :  3  :  6  ;  then,  by  alternation,  or  permutation,  it  will  be 
1  :  3  :  :  2  :  6. 

86.  Compounded  ratio  is,  when  the  sum  of  the  antecedent 
and  consequent  is  compared,  either  with  the  consequent,  or 
with  the  antecedent. — Thus,  if  1  :  2  :  :  3  :  6,  then,  by  composi- 
tion, 1  +  2  :  1  :  :  3  +  6  :  3,  and  1  4-  2  :  2  :  :  3  -f  6  :  6. 

87.  Divided  ratio,  is  when  the  difference  of  the  antecedent 
and  consequent  is  compared,  either  with  the  antecedent  or 
with  the  consequent. — Thus,  if  1  :  2  :  :  3  :  6,  then,  by  division, 
2-1:1  ::  0-3:  3,  and  2-1  :  2  :  :  6-3  :  6. 

Note.  The  term  Divided,  or  Division,  here  means  subtract- 
ing, or  parting  ;  being  used  in  the  sense  opposed  to  com- 
pounding, or  adding^  in  def.  86. 

THEOREM 


THEOREMS.  321 

THEOliEM  LXVI. 

Equimultiples  of  any  two  Quantities  have  the  same  Ratio  as 
the  Quantities  themselves. 
Let  a  and  b  be  any  two  quantities,  and  tma,  mB,  any 
equimultiples  of  them,  m  being  any  number  whatever  ;  thea 
will  mk  and  mB  have  the  same  ratio  as  a  and  b,  or  a  :  b  :  : 
mA  :  mB. 

mB        B 
For  —  =  — ,  the  same  ratio. 

WIA  A 

Corol.  Hence,  like  parts  of  quantities  have  the  same  ratio 
as  the  wholes  ;  because  the  wholes  are  equimultiples  of  the 
like  parts,  or  a  and  b  are  like  parts  of  mA  and  mB. 

THEOREM  LXVII. 

If  Four  Quantities,   of  the   Same  Kind,   be  Proportionals  ; 
they  will  be  in  Proportion  by  Alternation  or  Permutation, 
or  the   Antecedents  will  have  the  Same  Ratio  as  the  Con- 
sequents. 
Let  a   :  b   :  :  mA   :  mB  ;  then  will  a  :  mA   :  :  b   :  mB. 

mA  mB 

For  —  =  m,  and  —  =  m,  both  the  same  ratio. 

A  B 

THEOREM  LXVm. 

If  Four  Quantities  be  Proportional  ;    they  will  be  in  Pro- 
portion by  Inverlion,  or  Inversely. 
Let  a  :  b  :  :  mA  :  mB  ;  then  will  b   :  a  :  :  ms  :  mA. 

mA       A 
For  —  =  — ,  both  the  same  ratio. 
mB       B 

THEOREM  LXIX. 

If  Four  Quantities  be   Proportional  ;    they  will  be  ia  Pro- 
portion by  Composition  and  Division. 
Let  a  :  b   :  :  mA  :  mB  ; 
Then  will  b  ±:  a  :  a  :  :  mB  zt  mk  :  mA, 
and  B  ±  A  :  B  :  :  mB  it  mA  :  mB. 
mA  A  mB  B 

For, = ;  and = . 

mB  db  mA         b  ±  a  mB  rb  mA         ft  ±  a 

V*r..  L  42  CoroL 


32£  GEOMETRY. 

Corol.  It  appears  ffom  hence,  that  the  Sum  of  the  Greatest 
and  Least  of  four  proportional  quautities,  of  the  same  kind, 

exceeds  the  Sum  of  the   Two   Means.      For,  since 

A  :  A  -|-  B  :  :  mA  :  mA  -{-  mB,  where  a  is  the  least,  and 
wiA  -j-  »«B  the  greatest  ;  then  m  -f  1  •  a  +  wb,  the  sum  of 
the  greatest  and  least  exceeds  y/i  +  1  .  a  +  b  the  sum  of 
the  two  means. 

THEOREM  LXX. 

If,  of  Four  Proportional  Quantities,  there  betaken  any  Equi- 
multiples whatever  of  the  two  Antecedents,  and  any  Equi- 
multiples whatever  of  the  two  Consequents  ;  the  quantities 
resulting  will  still  be  proportional. 

Let  a  ;  b   :  :  mA  :  mB  ;    also,   let  pA  and  pmA    be    any 
equimultiples  of  the  two   antecedents,   and   ^b  and   ^mB  any 

equimultiples  of  the  two  consequents  ;    then  will 

jpA   :  9B   :  :  pmA  :  ^mB. 
qniB       qB 

For =  — ,  both  the  same  ratio. 

pmA     pA 

THEOREM  LXXI. 

If  there  be  Four  Proportional  Quantities,  and  the  two  Conse- 
quents be  either  Augmented  or  Diminished  by  Qjaantities 
that  have  the  Same  Ratio  as  the  respective  Antecedents  ; 
the  Results  and  the  Antecedents  will  still  be  Propor- 
tionals. 

Let  a  :  b  :  :  mA  :  jiiB,  and  ua  and  nmA  any  two    quan- 
tities having  the  same  ratio  as  the  two  antecedents  ;  then  will 
A   :   B  ±  nA   :  :  mA   :   mB  ±  nmA. 
mB  ■±L  nmA        b  ±:  nA 

For —  = ,  both  the  same  ratio. 

mA  A 

THEOREM  LXXn. 

If  any  Number    of  Quantities    be    Proportional,    then  any 
one  of  the  Antecedents  will  be  to  its  Consequent,  as  the 
Sum  of  all  the  Antecedents  is  to  the  Sum  of  all  the  Con- 
sequents. 
Let  A  :  b  :  :  mA  :  mB  :  :  ka  :  wb,&c.  ;  then  will 

A  :  B   :  :  a  -|-  mA  -j-  »a  :  :  b  -f  mB  +  «»>  &c. 

B  -f-  »"B  -f-  flB  b 

X    For "  =  — ,  the  same  ratio. 

a  +  mA  -{-  riA        A 

THEOREM 


THEOREMS.  3^3 


THEOREM  LXXin. 

U  a  Whole  Magnitude  be  to  a  Whole,  as  a  Part  taken  from 
the  first,  is  to  a  l^art  taken  from  the  other ;  then  the  Re- 
mainder will  be  to  the  Remainder,  as  the  whole  to  the 
whole. 

m  m 

Let  A  :  B  :  :  —  a  :  —  b  ; 


then  will  a:b::a a:b 1 

n  n 


-B 


B 


For     °     =  — ,  both  the  same  ratio. 

m 
A--A  A 

THEOREM  LXXIV. 

If  any  Qjuantities  be  Proportional ;  their  Squares,  or  Cubes, 
or  any  Like  Powers,  or  Roots,  of  them,  will  also  be  Pro- 
portional. 

Let  A  ;  b  :  :  mA  :  wiB  ;  then  will  a"  :  b"  :  :  m^A"  :  m"  b", 
m"B"        b" 

For =  — ,  both  the  same  ratio. 

w"a"       a"  ^ 

THEOREM  LXXV. 

If  there  be  two  Sets  of  Proportionals  ;  then  the  Products  or 
Rectangles  of  the  Corresponding  Terms  will  also  be  Pro- 
portional. 

Let  a  :  b  :  :  fOA  :  wb, 
and  c  :  D  : :  nc  :  no ; 
then  will  ac  :  bd  :  :  mnAC  :  mriBD. 

mriBD       bd 
For =  — ,  both  the  same  ratio. 

TnnAC       AC 

THEOREM  LXXVI. 

If  Four  Quantities  be  Proportional  ;  the  Rectanj^le  or  Product 
of  the  two  Extremes,  will  be   Equal  to  the  Rectangle  or 
Product  of  the  two  Means.     And  the  converse. 
Let  a  :  b  :  :  mA  :  niB  ; 
then  is  A  X  mB  =  b  X  mA  =  mAB,  as  is  evident. 

THEOREM 


324  GEOMETRY. 


THEOREM  LXXVII. 

If  Three  Quantities  be  Continued  Proportionals  ;  the  Rect- 
angle or  Product  of  the  tv.o  Extremes,  will  be  Equal  to  the 
Square  of  the  Mean.     And  the  converse. 
Let  a,  mA,  tw'^a  be  three  proportionals, 
or  A  :  mA  :  :  mA  :  m-A  ; 
then  is  A  X  m^A  =  m^AS,  as  is  evident. 


THEOREM  LXXVUI, 

If  any  Number  of  Quantities  be  Continued  Proportionals  j 
the  Ratio  of  the  First  to  the  Third,  will  be  duplicate  or  the 
Square  of  the  Ratio  of  the  First  and  Second  ;  and  the  Ratio 
of  the  First  and  Fourth  will  be  triplicate  or  the  cube  of 
that  of  the  First  and  Second  ;  and  so  on. 
Let  A,  mA,  m^A,  m^A,  &c.  be  proportionals  ; 
mA  m^A  m^A 

that  is  —  =  m  ;  but =  rn^  ;  and =  m^  ;  &c. 


THEOREM  LXXIX. 

Triangles,  and  also  Parallelograms,  having  equal  Altitudes, 
'  are  to  each  other  as  their  Bases. 

Let  the  two  triangles  ado,  def,  have  I  C  K. 

the  same  altitude,  or  between  the  same 
parallels  ae,  cf  ;  then  is  the  surface  of 
the  triangle  adc,  to  the  surface  of  the 
triangle  def,  as  the  base  ad  is  to  the  /  U  -X fi-ix  is^ 

base  DE.  Or,  ad  :  de  :  :  the  triangle 
ADC  :  the  triangle  def. 

For,  let  the  base  ad  be  to  the  base  de,  as  any  one  num- 
ber m  (2),  to  any  other  number  n  (3)  ;  and  divide  the  respec- 
tive bases  into  those  parts,  ab,  bd,  dg,  gh,  he,  all  equal 
to  one  another  ;  and  from  the  points  of  division  draw  the 
lines  Bc,  fg,  fh,  to  the  vertices  c  and  f.  Then  will 
these  lines  divide  the  triangles  adc,  def,  into  the  same  num- 
ber of  parts  as  their  bases,  each  equal  to  the  triangle 
ABC,  because  those  triangular  parts  have  equal  bases  and 
altitude  (corol.  '2,  th.  26)  ;  namely,  the  triangle  abc,  equal 
to  each  of  the  triangles  bdc,  dfg,  gfh,  hfe.  ^o  that 
the  triangle  adc,  is  to  the  triangle  dfe,  as  the  number  of 

parts 


THEOREMS.  326 

parts  m  (S)  of  the  former,  to  the  numher  n  (3)  of  the  latter, 
that  is,  as  the  base  ad  to  the  base  de  (def.  79) 

In  like  manner,  the  parallelogram  adki  is  to  the  parallelo- 
gram DEFK,  a«  the  base  ad  is  to  the  base  de  ;  each  of  these 
having  the  same  ratio  as  the  number  of  their  parts,  m  to  n. 

Q.    E.    D»- 

THEOREM  LXXX. 

Triangles,  and  also  Parallelograms,  having  Equal  Bases,  arc 
to  each  other  as  their  Altitudes. 

Let  ABC,  BEF,  be  two  triangles 
having  the  equal  basses  ab,  be,  and 
whose  altitudes  are  the  perpendicu- 
lars CG,  FH  ;  then  will  the  triangle 
ABC  :  the  triangle  bef  :  ;  cg   :  fh. 

For,  let  BK  be  perpendicular  to  ab, 
and  equal  to  cg  ;  in  which  let  there 
be  taken  bl=fh;  drawing  AK  and  AL. 

Then,  triangles  of  equal  base  and  heights  being  equal 
(corol.  2,  th.  25),  the  triangle  abk  is  =  abc,  and  the  triangle 
ABL  =  BEF  But  considering  now  abk,  abl,  as  two  triangles 
on  the  bases  bk,  bl,  and  having  the  same  altitude  ab,  these 
will  be  as  their  bases  (th.  79),  namely  the  triangle  abk  :  the 
triangle  abl   :  :  bk  :  bl. 

But  the  triangle  abk  =  abc,  and  the  triangle  abl  =  bef, 

also  bk  =   CG,  and  bl  =  fh. 
Theref.  the  triangle  abc  :  triangle  bef  :  :  cg  :  fh. 

And  since  parallelograms  are  the  doubles  of  these  triangles, 
having  the  same  bases  and  altitudes,  they  will  likewise  have 
to  each  other  the  same  ratio  as  their  altitudes.  q.  e.  d. 

Corol.  Since,  by  this  theorem,  triangles  and  parallelogram?, 
when  their  bases  are  equal,  are  to  each  other  as  their  ahi- 
tudes  ;  and  by  the  foregoing  one,  when  their  altitiKJes  are 
equal,  they  are  to  each  other  as  their  bases  ;  therefore  uni- 
versally, when  neither  are  equal,  they  are  to  ezth  other  in 
the  compound  ratio,'  or  as  the  rectangle  or  product  of  their 
bases  and  altitudes. 


THEOREM 


326  GEOMETRY. 


THEOREM  LXXXI. 


If  Four  Lines  be  Proportional  ;  the  Rectangle  of  the  Ex- 
tremes will  be  equal  to  the  Rectangle  of  the  Means. 
And,  conversely,  if  the  Rectangle  of  the  Extremes,  of  four 
Lines,  be  equal  to  the  Rectangle  of  the  Means,  the  Four 
Lines,  taken  alternately,  will  be  Proportional. 


Q 
B 

A 

P         D 

R 

Let  the  four  lines,  a,  b,  c,  d,  be  A 

proportionals,  or  a   :  b   :  :  c  :  d  ;  B 

then  will  the  rectangle  of  a  and  d  be  S. 

equal  to  the  rectangle  of  b   and  c  ; 
or  the  rectangle  a  .  d  =  b  .  c. 

For,  let  the  four  lines  be  placed 
with  their  four  extremities  meeting 
in  a   common  point,  forming  at  that 

point  four  right  angles  ;  and  draw  lines  parallel  to  them  to 
complete  the  rectangles  p,  q,  r,  where  p  is  the  rectangle  of  a 
and  D,  Q  the  rectangle  of  b  and  c,  and  r  the  rectangle  of  b 
and  D. 

Then  the  rectangles  p  and  r,  being  between  the  same 
parallels,  are  to  each  other  as  their  bases  a  and  b  (th,  79)  ; 
and  the  rectangles  q  and  r,  being  between  the  same  pa- 
rallels, are  to  each  other  as  their  bases  c  and  d.  But  the 
ratio  ef  a  to  b,  is  the  same  as  the  ratio  of  c  to  d  by  hypo- 
thesis ;  therefore  the  ratio  of  p  to  r,  is  the  same  as  the'ratio  of 
Q  to  R  ;  and  consequently  the  rectangles  p  and  ^  are  equal. 

Q.  E.    D. 

Again,  if  the  rectangle  of  a  and  d,  be  equal  to  the 
rectangle  of  b  and  c  ;  these  lines  will  be  proportional,  or 
A  :  B   :  :  c  ;  D. 

For,  the  rectangles  being  placed  the  same  as  before  :  then, 
because  parallelograms  between  the  same  parallels  are  to  one 
another  as  their  bases,  the  rectangle  p  :  r  :  :  a  :  b,  and 
q  :  R  :  :  c  ;  d.  But  as  p  and  q  are  equal,  by  supposition, 
they  have  the  same  ratio  to  r,  that  is,  the  ratio  of  a  to  b  is 
equal  to  the  ratio  of  c  to  d,  or  a  :  b  :  :  c  :  d.  ct.  e.  d. 

Corol.  1.  When  the  two  means,  namely,  the  second  and 
third  terms,  are  equal,  their  rectangle  becomes  a  square  of 
the  second  term,  which  supplies  the  place  of  both  the  second 
and  third.  And  hence  it  follows,  that  when  three  lines  are 
proportionals,  the  rectangle  of  the  two  extremes  is  equal  to 

the 


THEOREMS.  327 

I  the  square  of  the  mean  ;  and,  conversely,  if  the  rectangle  of 

R  the  extremes  be  equal  to  the  square  of  the  mean,  the  three 

f'  lines  are  proportionals. 

Corol.  2.  Since  it  appears,  bj  the  rules  of  proportion  in 
Arithmetic  and  Algebra,  that  when  four  quantities  are  propor- 
tional, the  product  of  the  extremes  ia  equal  to  the  product  of 
the  two  means  ;  and,  by  this  theorem,  the  rectangle  of  the 
extremes  is  equal  to  the  rectangle  of  the  two  means  ;  it  fol- 
lovvs,  that  the  area  or  space  of  a  rectangle  is  represented  or 
expressed  by  the  product  of  its  length  and  breadth  multiplied 
together.  And,  in  general,  a  rectangle  in  geometry  is  simi- 
lar to  the  product  of  the  measures  of  its  two  dimensions  of 
length  and  breadth,  or  base  and  height.  Also,  a  square  is  si- 
milar to,  or  represented  by,  the  measure  of  its  side  multiplied 
by  itself  So  that,  what  is  shown  of  such  products,  is  to  be 
understood  of  the  squares  and  rectangles. 

Corol.  3.  Since  the  same  reasoning,  ^s  in  this  theorem, 
holds  for  any  parallelograms  whatever,  as  well  as  for  the  rect- 
angles, the  same  property  belongs  to  all  kinds  of  parallelO'- 
grams,  having  equal  angles,  and  also  to  triangles,  which  are 
the  halves  of  parallelograms  ;  namely,  that  if  the  sides  about 
the  equal  angles  of  parallelograms  or  triangles,  be  reciprocally 
proportional,  the  parallelograms  or  triangles  will  be  equal  ; 
and,  conversely,  if  the  parallelograms  or  triangles  be  equal, 
their  sides  about  the  equal  angles  will  be  reciprocally  propor- 
tionali 

Coi-ol.  4.  Parallelograms,  or  triangles,  having  an  angle  in 
each  equal,  are  in  proportion  to  each  other  as  the  rectangles 
of  the  sides  which  are  about  these  equal  angles. 

THEOREM  LXXXIL 

If  a  Line  be  drawn  in  a  Triangle  Parallel  to  one  of  its  sides, 
it  will  cut  the  other  Sides  Proportionally. 

Let  de  be  parallel  to  the  side  bc  of  the 
triangle  abc  ;  then  will  ad  :  db  :  :  ae  ;  eg. 

For  draw  be  and  cb.  Then  the  tri- 
angles DBE,  DOE,  are  equal  to  each  other, 
because  they  have  the  same  base  de,  and 
are  between  the  same  parallels  de,  bc 
(th  25).  But  the  two  triangles  ade,  bde, 
on  the  bases  ad,  db,  have  the  same  alti- 

tilde  ; 


328  GEOMETRY. 

tude  ;  and  the  two  triangles  a©e,  cde,  on 
the  bases  ae,  ec,  have  also  the  same  alti- 
tude ;  and  because  triangles  of  the  same 
altitude  are  to  each  other  as  their  bases, 
therefore 

the  triangle  ade  :  bde  :  :  ad  :  db, 

and  triangle  ade  :  cde   :  :  ae  :  ec. 

But  BDE  is  =  CDE  ;  and  equals  must  have  to  equals  the  same 
ratio  ;   therefore  ad  :  db  :  :  ae  :  ec.  q.  e.  d. 

Corol.  Hence,  also,  the  whole  lines  ab,  ac,  are  proportional 
to  their  corresponding  proportional  segments  (corol.  th.  66). 


VIZ.  AB   :  AC   :  :  ad   :  ae, 
and  AB   :  AC   :  :   bd   :   ce. 


THEOREM  LXXXm. 

A  Line  which  Bisects  any  Angle  of  a  Triangle,  divides  the 
opposite  Side  into  Two  Segments,  which  are  Proportional 
to  the  two  other  Adjacent  Sides. 


E 


Let  the  angle  acb,  of  the  triangle  abc,  be 
bisected  by  the  line  en,  making  the  angle  r  >, 

equal  to    the  angle  s  :  then  will   the  seg-  ^  / 

ment  ad  be  to  the  segment  db.  as  the  side 
AC  is  to  the  side  cb.     Or,  ad  :  db  :  :  ac  :  cb. 


For,  let  be  be  parallel  to  cd,  meeting 
AC  produced  at  e.  Then,  because  the  line  bc  cuts  the  two 
parallels  cd,  be,  it  makes  the  angle  cbe  equal  to  thte  alternate 
angle  s  (th.  12),  and  therefore  also  equal  to  the  angle  r,  which 
is  equal  to  s  by  the  supposition.  Again,  because  the  line  ae  - 
cuts  the  two  parallels  dc,  be,  it  makes  the  angle  e  equal  to  the 
angle  r  on  the  same  side  of  it  (th.  14).  Hence,  in  the  triangle 
bce,  the  angles  b  and  e,  being  each  equal  to  the  angle  r,  are 
equal  to  each  other,  and  consequently  their  opposite  sides  cb, 
ce,  are  also  equal  (th.  3). 

But  now,  in  the  triangle  abe,  the  lime  cd,  being  draw^ 
parallel  to  the  side  be,  cots  the  two  other  sides  ab,  ae  propor- 
tionally (th.  82),  making  ad  to  db,  as  is  ac  to  ce  or  to  its 
equal  cb.  a.  e   d. 

THEOREM 


THEOREMS. 


329 


THEOREM  LXXXIV. 

Equiangular   Triangles  are  Similar,  or  ha?e  their  Like  Sides 
Proportional. 

Let  ABC,  DEF,  be  two  equiangular  tri-  q 

angles,  having  the  angle  a  equal  to  the 
angle  d,  the  angle  b  to  the  angle  e,  and 
consequently  the  angle  c  to  the  angle  f  ; 
then  will  AE  :  AC   :  :   de  :  df. 

For,  make  dg  =  ab,  and  dh  =  ac, 
and  join  gh.  Then  the  two  triangles 
ABC,  DGH,  having  the  two  sides  ab,  ac, 
equal  to  the  two  do,  dh,  and  the  con- 
tained angles  a  and  d  also  equal,  are  iden- 
tical, or  equal  in  all  respects  (th.  l),name- 
ly  the  angles  b   and   c  are  equal  to  the 

angles  g  and  h.  But  the  angles  b  and  c  are  equal  to  the  angles 
E  and  F  by  the  hypothesis  ;  therefore  also  the  angles  g  and  h 
are  equal  to  the  angles  e  and  f  (ax  I),  and  consequently  the 
line  GH  is  parallel  to  the  side  ef  (cor.  1,  th.  14). 

Hence  then,  in  the  triangle  def,  the  line  gh,  being  parallel 
to  the  side  ef,  divides  the  two  other  sides  proportionally, 
making  dg  :  dh  :  :  de  :  df  (cor.  th.  82).  But  dg  and  «h  are 
equal  to  ab  and  ac  ;  therefore  also  ab  :  ac  :  :  de  :  df. 


q,.  E.  D. 


THEOREM  LXXXV. 

Triangles  which  have  their  Sides  Proportional,  are  Equi- 
angular.^ 

In  the  two  triangles  abc,  def,  if 
ab  :  :  de  :  :  AC  :  df  :  :  bc  :  ef  ;  the  two 
triangles  will  have  their  corresponding 
angles  equal. 

For,  if  the  triangle  abc  be  not  equian- 
gular with  the  triangle  def,  suppose  some 
other  triangle,  as  deg,  to  be  equiangular 
with  ABC.  But  this  is  impossible  :  for  if 
the  two  triangles  abc,  deg,  were  equian- 
gular, their  sides  would  be  proportional 
(th.  84).  So  that,  ab  being  to  de  as  ac 
to  dg,  and  ab  to  np  as  eg  to  eg,  it  follows  that  dg  and  e» 
Wmg  fourth    proportionals    to  the    same  three    quantities 

Von.  I.  •  43  a9 


33U 


GEOMETRY. 


as  well  as  the  two  df,  ef,  the  former  dg,  eg,  would  be  equal 
to  the  latter,  df,  ef.  Thus  then,  the  two  trianjj^les,  def,  deg, 
having  t^ieir  three  sides  equal,  would  be  identical  (th.  6)  : 
which  is  absurd,  since  their  angles  are  unequal. 

THEOREM  LXXXVL 

Triangles,  which  have  an  Angle  in  the  one  Equal  to  an  Angle 
in  the  other,  and  the  Sitles  about  these  angles  Proportional, 
are  Equiangular. 

Let  ABC,  DEF,  be  two  triangles,  having 
the  angle  a  —  the  angle  d,  and  the  sides 
AB,  AC,  proportional  to  the  sides  de,  df  : 
then  will  the  triangle  abc  be  equiangular 
with  the  triangle  def. 

For,  make  dg  =  ab,  and  dh  =  ac, 
and  join  gh. 

Then,  the  two  triangles  abc,  dgh,  hav- 
ing two  sides  equal,  and  the  contamed 
angles  a,  and  d  equal,  are  identical  and 
equiangular  (th.  1),  having  the  angles  g 
and  H  equal  to  the  angles  b  and  c.  But,  since  the  sides  dg, 
DH,  are  proportional  to  the  sides  de,  df,  the  line  gh  is  parallel 
to  ef  (th.  82)  ;  hence  the  angles  e  and  f  are  equal  to  the  an- 
gles G  and  H  (th.  14),  and  consequently  to  their  equals  e  and  c. 


G  E 


THEOllEM  LXXXVIL 

in  a  Right- Angled  Triangle,  a  Perpendicular  from  the  Right 
Angle,  is  a  Mean  Proportional  between  the  Segments  of 
the  Hypothenuse  ;  and  each  of  the  Sides,  about  the  Right 
Angle,  is  a  Mean  Proportional  between  the  Hypothenuse 
and  the  adjacent  segment, 

^,.   .G 

Let  abc  be  a  right-angled  triangle,  and 
CD  a  perpendicular  from  the  right  angle 
c  to  the  hypothenuse  ab  ;  then  will 

CD  be  a  mean  proportional  between  ad  and  db  ; 

AC  a  mean  proportional  between  ab  and  ad  ; 

bo  a  mean  proportional  betwen  ab  and  bd. 

Or,  ad  :  CD   :  :  CD  :  db  ;  and   ab   :  bc  :  :  bc  :  bd  ;  and 

ab  :  AC   :  :  ac. 


For. 


THEOREMS. 


331 


For,  the  two  trian?;les  abc,  adc,  having  the  right  angijes 
at  c  an(J  d  equal,  and  the  angle  a  common,  have  their  third 
angles  equal,  and  are  equiangular  (corol.  I,  th.  17).  In  like 
manner,  the  two  triangles  abc,  bdc,  having  the  ri^ht  angles 
at  c  and  d  equal,  and  the  angle  b  common,  have  the  third  an- 
gles equal,  and  are  equiangular. 

Hence  then,  all  the  three  triangles  abc,  adc,  bdc, 
l?eing  equiangular  will  have  their  like  sides  proportional 
(th.  84;. 


VIZ.  AD  ;  CD  : 
and  AB  :  AC  : 
and  AB   :   bc   : 


:  CD  :  DB  ; 
;  AC  :  AD  ; 
:   BC   :   BD. 


Corol,  Because  the  angle  in  a  semicircle  is  a  right  angle 
(th.  52)  ;  it  follows,  that  if,  from  any  point  c  in  the  periphery 
of  the  semicircle,  a  perpendicular  be  drawn  to  the  diameter 
AB  ;  and  the  two  chords  ca,  cb,  be  drawn  to  the  extremities  of 
the  diameter  :  then  are  ac,  bc,  cd,  the  mean  proportionals  as 
in  this  theorem,  or  (by  th.  77),  cd^  =  ad  .  db  ;  ac^  =  ab  .  ad  ; 

and  BC2   =  AB  .  BD. 


THEOREM  LXXXVIIl. 

Equiangular    or  Similar  Triangles,  are  to  each  other  as  the 
Squares  of  their  Like  Sides. 


Let  ABC,  DEF,  be  two  equi- 
angular triangles,  ab  and  de 
being  two  like  sides  ;  then  will 
the  triangle  abc  be  to  the  tri- 
angle DEF,  as  the  square  of  ab 
is  to  the  square   of  de,    or  as 

AB^    to  DE^. 

For,  let   AL  and    dn    bc  the 


y 


B    D 


K 


squares  on  ab  and  de  ;  also  draw  their  diagonals  bk,  ev,  and 
the  perpendiculars  cg,  fh,  of  the  two  triangles. 

Then,  since  equiangular  triangles  have  their  like  sides 
proportional  (th.  84),  in  the  two  equiangular  triangles  abc, 
DEF,  the  side  ac  :  df  :  :  ab  :  de.;  and  in  the  two  acg, 
DFH,  the  side  ac  :  df  :  :  cg  :  fh  ;  therefore,  by  equaUty 
CG   :   FH   :  :   ab   :   de,  or  cg    :  ab   :  :   fh   :  de. 

But  because  triangles  on  equal  bases  are  to  each  other  as 
their  altitudes,  the  triangles  abc,  abk,  on  the  same  base 
AB,    are    to    each   other,    as  their  altitudes  cg,  ak,  or  ab  ; 

and 


332  OEOMETR?. 

and   the  triangles  dkf,  dem,  on  the  same  base  de,  are  as  theic 
altitudes  fh,  dm,  or  de. 

that  is.  triangle  abc  :  triangle  abk  :  :  cg  :  ab, 
and  triangle  def  :  triangle  dem  :  :  fh  :  de. 

But  it  has  been  shown  that  cg  :  ab  :  :  fh  :  de  ; 

theref.  of  equality  A  abc  :  A  abk  :  :  A  def  :  A  dem, 

er  alternately,  as  A  abc  :  A  def  :  :    A  abk  :  A  dem. 

But  the  squares  al,  dn,  being  the  double  of  tiie  triangleu 
ABH,  DEM,  have  the  same  ratio  with  them  ; 

therefore  the  A  abc  :   A  def   :  :  square  al   :  square  dn. 

ft.    E.    D. 
THEOREM  LXXXIX. 

All  Similar  Figures  are  to  each  other,  as  the  Squares  of  their 
Like  Sides. 

Let  abcde,  fghik,  be 
any  two  similar  figures,  the 
like  sides?  being  ab,  fg,  and 
bc,  gh,  and  so  on  in  the  same 
order  :  then  will  the  figure 
ABtDE  be  to  the  figure  fghik, 
as  the  square  of  ab  to  the 
square  of  fg,  or  as  ab»  to  rc^. 

For,  draw  be,  bd,  gk,  gi,  dividing  the  figures  into  asi 
equal  number  of  triangles,  by  lines  from  two  equal  angles 
b  and  G. 

The  two  figures  being  similar  (by  suppos.),  they,  are  equi- 
angular, and  have  their  like  sides  proportional  (def  70). 

Then,  since  the  angle  a  is  =  the  angle  f,  and  the  sidee 
AB,  ae,  proportional  to  the  sides  fg,  fk,  the  triangles 
ABE,  fgk,  are  equiangular  (th.  86).  In  like  manner,  the 
two  triangles  bcd,  ghi,  having  the  angle  c  =  the  angle  h, 
and  the  sides  bc,  cd,  proportional  to  the  sides  gh,  hi,  are 
also  equiangular  Also,  if  from  the  equal  angles  aed,  fki, 
there  be  taken  the  equal  angles  aeb,  fkg,  there  will  remain 
the  equals  bed,  gki  ;  and  it  from  the  equal  angles  cdc, 
hik,  be  taken  away  the  equals  cdb,  hig,  there  will  remain 
the  equals  bde,  gik  ;  so  that  the  two  triangles  bde,  gik, 
having  two  angles  equal,  are  also  equiangular.  Hence  each 
triangle  of  the  one  figure,  is  equiangular  with  each  corres- 
ponding triangle  of  the  other. 

But  equiangular  triangles  are  similar,  and  are  proportional 
to  the  squares  of  their  hke  sides  (th.  88). 

Therefore 


THEOREM!?. 


333 


Therefore  the  A  abe  :  A 
and  A  bcd  :  A 
and  A  bde  :   A 


FGK  :  :  ab2  :  fg«, 
GHi  :  :  Bc2  ;  GH^  ; 
GiK    :  :  de2   :  ik^. 


But  as  the  two  polygons  are  similar,  their  hke  sides  are  pro- 
portional, and  consequently  their  squares  also  proportional  ; 
so  that  all  the  ratios,  ab^  to  fg2,  and  bc3  to  gh2  and  de2  to 
ik2,  are  equal  among  themselves,  and  consequently  the  cor- 
responding triangles  also,  abe  to  fgk,  and  bcd  to  ghi,  and 
BDE  to  GIK,  have  all  the  same  ratio,  viz.  that  of  ab^  to  fg^  : 
and  hence  all  the  antecedents,  or  the  figure  abode,  have  to 
all  the  consequents,  or  the  figure  fghik,  still  the  same  ratio, 
viz.  that  of  ab2  to  fg2  (th.  72;.  Q.  e.  c 


THEOREM  XC. 


"Jimilar  Figures  Inscribed  ia  Circles,  have  their  Like  Sides, 
and  also  their  Whole  Perimeters,  in  the  Same  Ratio  as  the 
Diameters  of  the  Circles  in  which  they  are  Inscribed. 

Let  abode,  fghik, 
be  two  similar  figures, 
inscribed  in  the  circles 
whose  diameters  are  al 
and  FM  :  then  will  each 
side  ab,  bc,  &.c.  of  the 
one  figure  be  to  the  like 
side  GF,  GH,  &-C.  of  the 
other  figure,  or  the  whole  perimeter  ab  -{-  bc  -}-  &c.  of  the 
one  figure,  to  the  whole  perimeter  fg  -}-  gh  -f-  kc.  of  the 
other  figure,  as  the  diameter  al  to  the  diameter  fm. 

For,  draw  the  two  corresponding  diagonals,  ac,  fh,  as 
also  the  lines  bl,  gm.  Then,  since  the  polygons  are  similar, 
they  are  equiangular,  and  their  like  sides  have  the  same  ratio 
(def.  70)  ;  therefore  the  two  triangles  abc,  fgh,  have  the 
angle  b  =  the  angle  g,  and  the  sides  ab,  bc,  proportional 
to  the  two  sides  fg,  gh,  consequently  these  two  triangles 
are  equiangular  (th.  86),  and  have  the  angle  acb  =  fhg. 
But  the  angle  acb  =  alb,  standing  on  the  same  arc  ab  ; 
and  the  angle  fhg  =  fmg,  standing  on  the  same  arc  fg  ; 
therefore  the  angle  alb  =  fmg  (th.  1).  And  since  the 
angle  abl  =  fgm,  being  both  right  angles,  because  in  a 
semicircle  ;  therefore  the  two  triangles,  abl,  fgm,  having 
two  angles  equal,  are  equiangular  ;  and  consequently  their 
^  like 


334  ©EOMETRY. 

like  sides  are  proportional  (th.  84)  ;  hence  ab  :  fg   :  :  the 
diameter  al   :  the  diaraeter  fm. 

in  like  manner,  each  side  bc,  gd,  &c.  has  to  each  side 
GH,  HI,  &c.  the  same  ratio  of  al  to  fm  :  and  consequently 
th^  sums  of  them  are  still  in  the  same,  ratio  ;  viz.  ab  -|-  b<^  4* 
CD,  &c.  :  FG  -f-  GH  -|-   HI,  &c.  :  :  the  diam.  al  :  the  diam. 

FM  (th.  72).  ft.   E.    D. 

THEOREM  XCI. 

Similar  Figures  Inscribed  in  Circles,  are  to   each  other  as 
the  Squares  of  the  Diameters  of  those  Circles. 

Let  abcde,  fghik, 
be  two  similar  figures  in- 
scribed in  the  circles 
whose  diameters  are  al 
and  fm  ;  then  the  surface 
of  the  polygon  abode 
will  be  to  the  surface  of 
the  polygon  fghik,  as  al^  to  fm^  .  ^ 

For,  the  figures  being  similar,  are  to  each  other  as  the 
squares  of  their  like  sides,  ab^  to  fg^  (th.  88).  But.  by 
the  last  theorem,  the  sides  ab,  fg,  are  as  the  diameters  al, 
FM  ;  and  therefore  the  squares  of  the  sides  ab^  to  rc^ ,  as  the 
squares  of  the  diameters  al2  to  fm^  (th.  74).  Consequently 
the  polygons  abcde,  fghik,  are  also  to  each  other  as  the 
squares  of  the  diameters  al^  to  fm^  (ax.  1).  Q.  e    d.> 


THEOREM  XCH. 

The  Circumferences  of  all  Circles  are  to  each  other-  as  their 
Diameters. 

Let  D,  d,  denote  the  diameters  of  two  circles,  and  c,  c, 
their  circumferences  ; 

then  will  d  :  </  :  :  c  :  c,  or  d  :  c   :  :  d  :  c. 

For,  (by  theor.  90,)  sinjilar  polygons  inscribed  in  circles 
have  their  perimeters,  in  the  same  ratio  as  the  diameters  of 
those  circles. 

Now,  as  this  property  belongs  to  all  polygons,  whatever 
the  number  of  the  sides  may  be  ;  conceive  the  number  of  the 
sides  to  b«:  indefinitely  great,  and  the  length  of  each  inde- 
finitely small,  till  they  coincide  with  the  circumference  of 

the 


THEOREMS.  335 

the  circle,  and  be  equal  to  it,  indefinitely  neap.  Thee  tbe 
perimeter  of  the  poly^^on  of  an  infinite  number  of  sides,  is 
the  same  thing  as  the  circumference  of  the  circle.  Hence  it 
appears  that  the  circumferences  of  the  circles,  being  the  same 
as  the  perimeters  of  such  polygons,  are  to  each  other  in  the 
same  ratio  as  the  diameters  of  the  circles.  q.  e.  d. 

THEOREM  XCni. 

The  Areas  or  Spaces  of  Circles,  are  to  each  other  as  the 
Squares  of  their  Diameters,  or  of  their  Radii. 

Let  a,  a,  denote  the  areas  or  spaces  of  two  circles,  and  d, 
d  their  diameters  ;  then  a  :  a  :  :  d^  :  d^ . 

For  (by  theorem  91)  similar  polygons  inscribed  in  circles 
are  to  each  other  as  the  squares  of  the  diameters  of  the  cir- 
cles. 

Hence,  conceiving  the  number  of  the  sides  of  the  polygons 
to  be  increased  more  and  more,  or  the  length  of  the  sides  to 
become  less  and  less,  the  polygon  approaches  nearer  and  near- 
er to  the  circle,  till  at  length,  by  an  infinite  approach,  coin- 
cide, and  become  in  effect  equal  ;  and  then  it  follows  that  the 
spvices  of  the  circles,  which  are  the  same  as  of  the  polygons, 
will  be  to  each  other  as  the  squares  of  the  diameters  of  the 
circles.  q.  e.  b. 

Coral,  The  spaces  of  circles  are  also  to  each  other  as  the 
squares  of  the  circumferences  ;  since  the  circumferences  are 
in  the  same  ratio  as  the  diameters  (by  theorem  92). 

THEOREM  XCIV. 

The  Area  of  any  Circle,  is  Equal  to  the  Rectangle  of  Half 
its  Circumference  and  half  its  Diameter. 

Conceive  a  regular  polygon  to  be  in-  ^ — ^ 

scribed  in  the  circle  :  and  radii  drawn  to  /f\  \   /\\ 

all  the  angular  points,  dividing  it  into  as  //  ^   \:/     \\ 

many  equal  triangles  as  the  polygon  has  r — ~Q^v ^ 

sides,  one  of  which  abc,  of  which   the  Vv    /  :\    // 

altitude  is  the  perpendicular  cd  from  the  \/   \   \// 

centre  to  the  base  ae.  j^T^SP^ 

Then  the  triangle  abc,  being  equal  to 
a  rectangle  of  half  the  base  and  equal  altitude  (th.  26,  cor.  2), 
is  equal  to  the  rectangle  of  the  half  base  ad  and  the  altitude  co  ; 


336  GEOMETRY. 

consequently  the  whole  polygon,  or  all 
the  triangles  added  together  which  com- 
pose it,  is  equal  to  the  rectangle  of  the 
common  altitude  cd.  and  the  halves  of  all 
the  sides,  or  the  half  perimeter  of  the 
polygon. 

Now,  conceive  the  number  of  sides  of  the  polygon  to  be 
indefinitely  increased  ;  then  will  its  perimeter  coincide  with 
the  circumference  of  the  circle,  and  consequently  the  altitude 
CD  will  become  equal  to  the  radius,  and  the  whole  polygon 
equal  to  the  circle.  Consequently  the  space  of  the  circle,  or 
of  the  polygon  in  that  state,  is  equal  to  the  rectangle  of  the 
radius  and  half  the  circumference.  q.  e.  b. 


OF  PLANES  AND  SOLIDS. 


DEFINITIONS. 

Def.  88.  The  common  Section  of  two  Planes,  is  the  line 
in  which  they  meet,  to  cut  each  other. 

89.  A  Line  is  Perpendicular  to  a  Plane,  when  it  is  per- 
pendicular to  every  line  in  that  plane  which  meets  it. 

90.  One  Plane  is  Perpendicular  to  Another,  when  every 
line  of  the  one,  which  is  perpendicular  to  the  line  of  their 
common  section,  is  perpendicular  to  the  other. 

91.  The  inclination  of  one  Plane  to  another,  or  the  angle 
they  form  between  them,  is  the  angle  contained  by  two 
lines  drawn  from  any  point  in  the  common  section,  and  at 
right  angles  to  the  same,  one  of  these  lines  in  each  plane. 

92.  Parallel  Planes,  are  such  as  being  produced  ever  so 
far  both  ways,  will  never  meet,  or  which  are  every  where  at 
an  equal  perpendicular  distance. 

93.  A  Solid  Angle,  is  that  which  is  made  by  three  or  more 
plane  angles,  meeting  each  otlier  in  the  same  point. 

94.  Similar 


DEFINITIONS. 


337 


\          N 

1 

^         ^ 

94.  Similar  Solids,  contained  by  plane  figures,  are  such  as 
have  all  their  solid  angles  equal,  each  to  each,  and  are  bound- 
ed by  the  same  number  of  similar  planes,  alike  placed. 

95.  A  Prism,  is  a  solid  whose  ends  are  parallel,  equal,  and 
like  plane  figures,  and  its  sides,  connecting  those  ends,  are 
parallelograms. 

96.  A  Prism  takes  particular  names  according  to  the  figure 
of  its  base  or  ends,  whether  triangular,  square,  rectangular, 
pentagonal,  hexagonal,  &.c^ 

97.  A  Right  or  Upright  Prism,  is  that  which  has  the 
planes  of  the  sides  perpendicular  to  the  planes  of  the  ends 
or  base. 

98.  A  Parallelopiped,  or  Parallelopipedon,  is 
a  prism  bounded  by  six  parallelograms,  every 
opposite  two  of  which  are  equal,  alike,  and  pa- 
rallel. 

99.  A  Rectangular  Parallelopipedon,  is  that  whose  bound- 
ing planes  are  all  rectangles,  which  are  perpendicular  to  each 
other. 

100.  A  Cube,  is  a  square  prism,  being  bound- 
ed by  six  equal  square  sides  or  faces,  and  are 
perpendicular  to  each  other. 

101.  A  Cylinder,  is  a  round  prism,  having  cir- 
cles for  its  ends  ;  and  is  conceived  to  be  formed 
by  the  rotation  of  a  right  line  about  the  circum- 
ferences of  two  equal  and  parallel  circles,  always 
parallel  to  the  axis. 

102.  The  Axis  of  a  Cylinder,  is  the  right  line 
joining  the  centres  of  the  two  parallel  circles,  about  which 
the  figure  is  described. 

103.  A  Pyramid,  is  a  solid,  whose  base  is  any 
right-lined  plane  figure,  and  its  sides  triangles, 
having  all  their  vertices  meeting  together  in  a 
point  above  the  base,  called  the  Vertex  of  the 
pyramid. 

104.  A  pyramid,  like  the  prism,  takes  particular  names 
from  the  figure  of  the  base. 

105.  A  Cone,  is  a  round  pyramid,  having  a  cir-  A 
cular  base,   and  is  conceived  to    be  generated  by  /  \ 
the  rotation  of  a  right  line  about  the  circumference  L--\ 
of  a  circle,  one   end  of  which  is  fixed  at  a  point  CJ' 
above  the  plane  of  that  circle. 


J^ 

7 

A 

^ 

Vol.  I. 


44 


19S.  The 


338  GEOMETRY. 

106.  The  Axis  of  a  cone,  is  the  right  line,  joining  the 
vertex,  or  fixed  pQint,  and  the  centre  of  ilie  circle  about 
which  the  figure  is  described. 

107.  Similar  Cones  and  Cylinders,  are  such  as  have  their 
altitudes  and  the  diameters  of  their  bases  proportional. 

108.  A  Sphere,  is  a  solid  bounded  by  one  curve  surface, 
which  is  every  where  equally  distant  from  a  certain  point 
within,  called  the  Centre.  It  is  conceived  to  be  generated 
by  the  rotation  of  a  semicircle  about  its  diameter,  which  re- 
mains fixed. 

109.  The  Axis  of  a  Sphere,  is  the  right  line  about  which 
the  semicircle  revolves  ;  and  the  centre  is  the  same  as  that 
of  the  revolving  semicircle. 

110.  The  Diameter  of  a  Sphere,  is  any  right  line  passing 
through  the  centre,  and  terminated  both  ways  by  the  surface. 

111.  The  Altitude  of  a  Sohd,  is  the  perpendicular  drawn 
from  the  vertex  to  the  opposite  side  or  base. 

THEOREM  XCV. 

A  Perpendicular  is  the  Shortest  Line  which  can  be  draws 
from  any  Point  to  a  Plane. 

Let  ab  be  perpendicular  to  the  plane  . 

DE  ;  then  any  other  line,  as  ac,  drawn  \ 

from  the  same  point  a  to  the  plane,  will  \ 

be  longer  than  the  line  ab.  ^^^—a^.,,.^^ 

In  the  plane  draw  the  line  bc,  joining          dC   B — ^CyE 
the  points  B,  c.  

Then,  because  the  line  ab  is  perpendi- 
cular to  the  plane  de,  the  angle  b  is  a  right  angle  (def.  89), 
and  consequently  greater  than  the  angle   c  ;    therefor^  the 
line.AB  opposite  to  the  less  angle,  is  less  than  any  other  line 
AC,  opposite  the  greater  angle  (th.  21).  q,  e-  d. 

THEOREM  XCVI. 

A  Perpendicular  Measures  the  Distance  of  any  Point  from  a 
Plane. 

The  distance  of  one  point  from  another  is  measured  by  a 
right  line  joining  them,  because  this  is  the  shortest  line  which 
can  be  drawn  from  one  point  to  another.  So,  also,  the 
distance  from  a  point  to  a  line,  is  measured  by  a  perpendi- 
cular, because  this  line  is  the  shortest  which  can  be  drawn 

from 


THEOREMS. 


339 


from  the  point  to  the  line.  In  like  manner,  the  distance  from 
a  point  to  a  plane,  must  be  measured  by  a  perpendicular 
drawn  from  that  point  to  the  plane,  because  this  is  the  short- 
est line  which  can  be  drawn  from  the  point  to  the  plane. 


THEOREM  XCVII. 


The  common  Section  of  Two  Planes,  is  a  Right  Line. 


.\ 


A' 


\ 


1 


Let  acbda,  aebfa,  be  two  planes 
tfuttiogeach  pther,  and  a,  b,  two  points 
in  which  the  two  planes  meet :  drawing 
the  line  ab,  this  line  will  be  the  common 
intersection  of  the  two  planes. 

For  because  the  right  line  ab  touches 
the  two  planes  in  the  points  a  and  b,  it 
touches  them  in  all  other  points  (def. 
20)  :  this  hue  is  therefore  common  to  the  two  planes.     That 
is,  the  common  intersection  of  the  two  planes  is  a  right  line. 

Q.    E.    D. 


D 


THEOREM  XCVm. 


B 


If  a  Line  be  Perpendicular  to  two  other  Lines,  at  their  Com- 
mon  Point  of  Meeting ;  it  will   be    Perpendicular  to    the 
Plane  of  those  Lines. 
Let  the  line   ab   make    right  angles 

with  the  lines  ac,  ad  ;  then  will  it  be 

perpendicular   to  the   plane  cde  which 

passes  through  these  lines. 

If  the  line  ab  were  not  perpendicular 

to  the  plane  cde,  another  plane  misjht 

pass  through  the  point  a,  to  which  the 

line   ab  would  be  perpendicular.     But 

this  is  impossible  ;  for,  since   the  angles   bag,  bad,  are  right 

angles,  this  other  plane  must  pass  through  the  points  c,  d. 

Hence,  this  plane  passing  through  the  two  points  a,  c,  of  the 

line  AC,  and  through  the  two  points  a,  d,  of  the  line  ad,   it 

will  pass  through  both  these  two  lines,  and  therefore  be  the 

same  plane  with  the  former.  q.  e.  d. 


CS) 


THEOREM 


340 


eEOMETRY. 


THEOREM  XCIX. 


Tf  Two  Lines  be  Perpendicular  to  the  Same  Plane,  they  will 
be  Parallel  to  each  other. 

Let  the  two  lines  ab,  cd.  be  both  per- 
pendicular to  the  same  plane  ebdf  ;  then 
will  AB  be  parallel  to  cd. 

For,  join  b,  d,  by  the  line  bd  in  the 
plane.  Then,  because  the  lines  ab,  cd, 
are  perpendicular  to  the  plain  ef,  they 
are  both  perpendicular  to  the  line  bd  (def.  89)  in  that  plane  ; 
and  consequently  they  are  parallel  to  each  other  (corol.  th. 
13).  Q.  e:   d*. 

Corol.  If  two  lines  be  parallel,  and  if  one  of  them  be 
perpendicular  to  any  plane,  the  other  will  also  be  perpendicu* 
far  to  the  same  plane. 


THEOREM  C, 


If  Two  planes  Cut  each  other  at  Right  Angles,  and  a  Line  be 
drawn  in  one  of  the  Planes  Perpendicular  to  their  Common 
Intersection,  it  will  be  Perpendicular  to  the  other  Plane. 


A^ 


^, 


Let  the  two  planes  acbd,  aebf,  cut 
each  other  at  right  angles  ;  and  the  line 
CG  be  perpendicular  to  their  common 
section  ab  ;  then  will  cg  be  also  perpen- 
dicular to  the  other  plane  aebf. 

For,  draw  eg   perpendicular   to  ab.  -q 

Then  because  the  two  lines  gc,  ge,  are 

perpendicular  to  the  common  intersection  ab,  the  angle  cge 
is  the  angle  of  incUnation  of  the  two  planes  (def.  91).  But 
since  the  two  phnes  cut  each  other  perpendicularly,  the 
angle  of  inclination  cge  is  a  right  angle.  And  since  the  line 
CG  is  perpendicular  to  the  two  lines  ga,  ge,  in  the  plane  aebf> 
it  is  therefore  perpendicular  to  that  plane  (th.  98). 

Q.  e.  d. 


*  This  demonstratimi  of  Theorem,  xcix.  does  not  appear  to  me' to 
fee  conclude..  EDI  roR.  .  THEOREM 


THEOREMS.  341 


THEOREM  CI. 


If  one  Plane  Meet  another  Plane,  it  will  make  angles  with 
that  other  Plane,  which  are  together  equal  to  two  Right 
Angles. 

Let  the  plane  acbc  meet  the  plane  aebf  ;  these  planes 
make  with  each  other  two  angles  whose  sum  is  equal  to  two 
right  angles. 

For,  through  any  point  Oj  in  the  common  section  ab,  dravr 
CD,  EF,  perpendicular  to  ab.  Then,  the  line  cg  makes  with 
EF  two  angles  together  equal  to  two  right  angles.  But  these 
two  angles  are  (by  def.  91)  the  angles  of  inclination  of  the  two 
planes.  Therefore  the  two  planes  make  angles  with  each 
other,  which  are  together  equal  to  two  right  angles. 

Corol.  In  like  manner  it  may  be  demonstrated,  that  planes 
which  intersect,  have  their  vertical  or  opposite  angles  equal  ; 
also,  that  parallel  planes  have  their  alternate  angles  equal  ; 
and  so  on,  as  in  parallel  lines. 


THEOREM  CU. 

If  Two  Planes  be  Parallel  to  each  other ;  a  Line  which  is 
Perpendicular  to  one  of  the  Planes,  will  also  be  Perpendi- 
cular to  the  other. 

Let  the  two  planes  cd,  ef,  be  parallel, 

and  let  the  line  ab  be  perpendicular  to  the  ^ — -^ 

plane  cd  :  then  shall  it  also  be  perpendicu-  ^f^ — 1  )^ 

lar  to  the  other  plane  ef.  ^J — r 

For,  from  any  point  g,  in  the  plane  ef, 
draw  GH  perpendicular  to  the  plane  cd,  and 
draw  . 


raw  ah,  bg,  ^ ^^ 

Then,  because  ba,  gh,  are  both  perpendi-  ^\A. — H/^ 

cular  to  the  plane  cd,  the  angles  a  and  h  are  

both  right  angles.  And  because  the  planes  cd,  ef,  are 
parallel,  the  perpendiculars  ba,  gh,  are  equal  (def  92). 
Hence  it  follows  that  the  lines  bg,  ah,  are  parallel  (^Aef,  9)* 
And  the  line  ab  being  perpendicular  to  the  line  ah,  is  also  per- 
pendicular to  the  parallel  line  bg  (cor.  th.  12). 

In  like  manner  it  is  proved,  that  the  line  ab  is  perpendi- 
cular to  all  other  lines  whick  can  be  drawn  from  the  point  b 

in 


342 


GEOMETRY. 


in  the  plane  ef.     Therefore  the  line  ab  is  perpendicular  to 
the  whole  plane  ef  (def.  92).  ^  *      ^.  e.  d. 


THEOREM  cm. 


If  Two  Lines  be  Parallel  to  a  Third  Line,  though  not  in  the 
same  Plane  with  it  ;  they  will  be  Parallel  to  each  other. 


F 


.^^Tx 


Let  the  lines  ab,  cd,  be  each  of  them 
parallel  to  the  third  line  ef,  though  not  in 
the  same  plane  with  it  ;  then  will  ab  be  pa-  "Q 

rallel  to  cd. 

For,  from  any  point  g  in  the  line  ef,  let  gh, 
Gi,  be  each  perpendicular  to  ef,  in  the  planes  H 

eb,  ED,  of  the  proposed  parallels.  . 

Then,  since  the  line  ef  is  perpendicular  /E\ 

to  the  two  lines  gh,  gi,  it  is  perpendicular  ^ 
to  the  plane  ghi  of  those  lines  (th  98).  And  because  ef  is 
perpendicular  to  the  plane  ghi,  its  parallel  ab  is  also  perpen- 
dicular to  that  plane  (cor.  th.  99.)  For  the  same  reason,  the 
line  CD  is  perpendicular  to  the  same  plane  ghi.  Hence,  be- 
cause the  two  hnes  ab,  cd,  are  perpendicular  to  the  same  plane 
these  two  lines  are  parallel  ^th.  99).  ^.  e.  d. 


THEOREM  CIV. 


If  Two  Lines,  that  meet  each  other,  be  Parallel  to  Two  other 
Lines  that  meet  each  other,  though  not  in  the  same  Plane 
with  them  ;  the  Angles  contained  by  those  Lines  will  be 
equal. 


Let  the  two  lines  ab,  bc,  be  parallel  to 
the  two  lines  de,  ef  ;  then  will  the  angle  abc 
be  equal  to  the  angle  def. 

For,  make  the  lines  ab,  bc,  de,  ef,  all 
equal  to  each  other,  and  join  ac,  df,  ad,  be, 
of.  . 

Then,  the  lines  ad,  be,  joining  the  equal 
and  parallel  lines  ab,  de,  are  equal  and  paral-  j^ 
lei  (th.  24).  For  the  same  reason,  cf,  be, 
are  equal  and  parallel.  Therefore  ad,  cf,  are  equal  and  pa- 
rallel (th.  15)  ;  and  consequently  also  ac,  DF(th.  24).  Hence 
the  two  triangles  abc,  def,  having  all  their  sides  equal, 
)  each 


kKl 


THEOREMS. 


343 


each  to  each,  have  their  angles  also  equal,  and  consequently 
the  angle  abc  =  the  angle  def.  q.  e.  d. 

THEOREM  CV. 


The  Sections  made  by  a  Plane  cutting  two  other  Parallel 
Planes,  are  also  Parallel  to  each  other. 

Let  the  two  parallel  planes  ab,  cd,  be 
out  by  the  third  plane  ekhg,  in  the  lines 
EF,  GH  :  these  two  sections  ef,  gh,  will 
be  parallel. 

Suppose  FG,  FH,  be  drawn  parallel  to 
Bach  other  in  the  plane  eehg  ;  also  let  ei, 
FK,  be  perpendicular  to  the  plane  cd  ;  and 
let  ig,  kh,  be  joined. 

Then  eg,  fh,  being  parallels,  and  ei,  fk,  hemg  both  per- 
pendicular to  the  plane  cd,  are  also  parallel  to  each  other 
(th.  99)  ;  consequently  the  angle  hfk  is  equal  to  the  angle 
GEi  (th.  104).  But  the  angle  fkh  is  also  equal  to  the  angle 
EiG,  being  both  right  angles  ;  therefore  the  two  triangles  are 
equiangular  (cor.  1,  th.  17)  ;  and  the  sides  fk,  ei,  being  the 
equal  distances  between  the  parallel  planes  (def.  92),  it  fol- 
lows that  the  sides  fh,  eg,  are  also  equal- (th.  2)  But  these 
two  hnes  are  parallel  (by  suppos.),  as  well  as  equal  ;  conse- 
quently the  two  lines  ef,  gh,  joining  those  equal  parallels,  are 
also  parallel  (th.  24).  Q.  e.  d. 


THEOREM  CVI. 


If  any  Prism  be  cut  by  a  Plane  Parallel  to  its  Base,  the  Sec- 
tion will  be  equal  and  Like  to  the  Base. 

Let  AG  be  any  prism,  and  il  a  plane  pa- 
rallel to  the  base  ac  ;  then  will  the  plane  il 
be  equal  and  like  to  the  base  ac,  or  the  two 
planes  will  have  all  their  sides  and  all  their 
angles  equal. 

For  the  two  planes  ac,  il,  being  parallel, 
by  hypothesis  ;  and  two  parallel  planes,  cut 
by   a  third  plane,    having  parallel    sections  ^  J3 

(th.  105)  ;  therefore  ik  is  parallel  to  ab,  and 
KL  to  Bc,  and  im  to  cd,  and  iiM  to  ad.     But  ai  and  bk  are 
parallels  (by  def.  95)  consequently  ak  is  a  parallelogram  ; 
and  the  opposite  sides  ab,  ik,  are  equal  (th.  22).     In  like 

manner, 


3 

b:   g 

344 


GEOMETRY. 


IrL 


.1 


G 


K 


B 


manner,  it  is  shown  that  kl  is  =  bc,  and  lm 
=  CD,  and  iM  =  AD,  or  the  two  planes  ac,  il, 
are  mutually  equilateral.  But  these  two 
planes,  having  their  corresponding  sides  pa- 
rallel, have  the  angles  contained  by  them  also 
equal  (th.  104),  namely,  the  angle  a  =  the 
angle  i,  the  angle  b  =  the  angle  k,  the  angle 
c  =  the  angle  l,  and  the  angle  d  =  the  angle 
M.  So  that  the  two  planes  ac,  il,  have  all 
their  corresponding  sides  and  angles  equal,  or  they  are  equal 
and  like.  >  q.  e.  d. 

THEOREM  CVII. 

If  a  Cylinder  be  cut  by  a  Plane  Parallel  to  its  Base,  the  Sec- 
tion will  be  a  Ci^-cle,  Equal  to  the  Base. 

Let  af  be  a  cylinder,  and  ghi  any 
section  parallel  to  the  base  abc  ;  then  will 
QHi,  be  a  circle  equal  to  abc 

For,  let  the  planes  ke,  kf  pass  through 
the  axis  of  the  cylinder  mk,  and  meet  the 
section  ghi  in  the  three  points  h,  i,  l  ; 
and  join  the  points  as  in  the  figure* 

Then,  since  kl,  cl,  are  parallel  (by  def. 
101)  ;  and  the  plane  ki  meeting  the  two 
parallel  planes  abc,  ghi,  makes  the  two  sectiens  kc,  li,  pa- 
rallel (th.  105)  ;  the  figure  klic  is  therefore  a  parallelogram, 
and  consequently  has  the  opposite  sides  li,  kc,  equal,  where 
Kc  is  a  radius  of  the  circular  base. 

In  like  manner  it  is  shown  that  lh  is  equal  to  the  radius 
KB  ;  and  that  any  other  lines,  drawn  from  the  point  l  to  the 
circumference  of  the  section  ghi,  are  all  equal  to  radii  of  the 
base  ;  consequently  ghi  is  a  circle,  and  equal  to  abc.     q.  e.  d. 

THEOREM  CVIII. 

All  Prisms  and  Cylinders,  of  equal  Bases  and  Altitudes,  are 
Equal  to  each  other. 


Let  AC,  DF,  be  two 
prisms,  and  a  cylinder, 
©n  equal  bases  ab,  de, 
and  having  equal  alti- 
tudes B«,  FF  ;  then  will 
the  solids  ac,  df,  be 
equal. 

Foxy  let  PQ,  Rs,   be' 
>     T,  .  any 


H 

/  / 

/  ) 

THEOREMS. 


345 


any  two  sections  parallel  to  the  bases,  and  equidistant  from 
them.  Then,  by  the  last  two  theorems,  the  section  pq  is 
equal  to  the  base  ab,  and  the  section  rs  equal  to  the  base 
DE.  But  the  bases  ab,  de,  are  equal,  by  the  hypothesis  ; 
therefore  the  sections  pq,  rs,  are  equal  also.  In  like  manner, 
it  may  be  shown,  that  any  other  corresponding  sections  are 
equal  to  one  another. 

bince  then  every  section  in  the  prism  ac,  is  equal  to  its 
corresponding  section  in  the  prism  or  cylinder  df,  the  prisms 
and  cylinder  themselves,  which  are  composed  of  an  equal 
number  or  all  those  equal  sections,  must  also  be  equal,  q..  e.  d. 

Carol.  Every  prism,  or  cylinder,  is  equal  to  a  rectangular 
parallelopipedon,  of  an  equal  base  and  altitude. 

THEOREM  CIX. 

Rectangular   Parallelopipedons,   of  Equal   Altitudes,  are   to 
each  other  as  their  Bases. 


\  ^ 

V      ^ 

^     \ 

D 
I 

K 

B 

\     ^ 

\     ^ 

\     \ 

V  ^ 

a' 

0 

Let   AC,   EG,  be  two   fectan- 
gular   parallelopipedons,  having    Q    R    S  ^C  T  "V    G 

the  equal  altitudes  ad,  eh  ; 
then  will  the  solid  ac  be  to  the 
solid  E6,  as  the  base  ab  is  to 
the  base  ef. 

For,  let  the  proportion  of  the 

base  AB  to  the  base  ef,  be  that    ^^    I^    M   K  E  P 

of  any  one  number  m  (3)  to  any 

other  number  n  (2).  And  conceive  ab  to  be  divided  into  m 
equal  parts,  or  rectangles,  ai,  lk,  mb,  (by  dividing  an  into 
that  number  of  equal  parts,  and  drawing  il,  km,  parallel 
to  bn).  And  let  ef  be  divided,  in  like  manner,  into  n  equal 
parts,  or  rectangles,  eo,  pf  :  all  of  these  parts  of  both  bases 
being  mutually  equal  among  themselves.  And  through  the 
lines  of  division  let  the  plane  sections  lr,  ms,  fv,  pass  parallel 

to  AQ,  ET. 

Then,  the  parallelopipedons  ar,  ls,  mc,  ev,  pg,  are  all 
equal,  having  equal  bases  and  altitudes.  Therefore  the  soHd 
AC  is  to  the  solid  eg,  as  the  number  of  parts  in  the  former, 
to  the  number  of  equal  parts  in  the  latter  ;  or  as  the  number 
of  parts  in  ab  to  the  number  of  equal  parts  in  ef,  that  is,  as 
the  base  ab  to  the  base  ef.  q.  e.  d. 

Corol.    From  this  theorem,  and  the  corollary  to  the  last,  it 

appears,  that  all  prisms  and  cylinders  of  equal  altitudes,  are 

Vo*.  1.  45  t« 


346 


<IEOMBTRY. 


to  each  other  as  their  bases  ;  every  prism  and  cylinder  being 
equal  to  a  rectangular  pirallelopipedon  of  an  equal  base  and 
altitude. 

tHEOREM  ex. 

Rectaiigular  Parallelopipedons,  of  Equal  Bases,  are  to  each 
other  as  their  Altitudes. 

Let  ab,  cd,  be  two  rectan- 
gular parallelopipedons,  stand- 
ing on  the  equal  bases  ae,  cf  ; 
then  will  the  solid  ab  be  to  the* 
solid  CD,  as  the  altitude  eb  is  to 
the  altitude  fb. 

For,  let  AG  be  a  rectangular 
parallelopipedon  on  the  base  ae, 
and  its  aUitude  eg  equal  to  the  altitude  fd  of  the  solid  cd. 

Then  ag  and  cd  are  equal,  being  piisnos  of  equal  bases 
and  altitudes.  But  if  hb,  hg,  be  considered  as  bases,  the 
solids  ab,  AG,  of  equal  altitude  ah,  will  be  to  each  other 
as  those  bases  hb,  hg.  But  these  bases  hb,  hg,  being 
parallelograms  of  equal  altitude  he,  are  to  each  other  as 
.their  bases  eb,  eg  ;  therefore  the  two  prisms  ab,  ag,  are 
to  each  other  as  the  Hues  eb,  eg.  But  ag  is  equal  to 
CD,  and  eg  equal  to  fd  ;  consequently  the  prisms  ac,  cd, 
are  to  each  other  as  their  altitudes  eb,  fd  ;  that  is,  -  -  - 
AB   :  CD  :  :  eb   :   fd.  q.  e.  d. 

CoroL  1.  Ftom  this  theorem,  and  the  corollary  to  theorem 
1G8,  it  appears,  that  all  prisms  and  cylinders,  of  equal  bases, 
are  to  one  another  as  their  altitudes. 

Carol.  2.  Because,  by  corollary  1,  prisms  and  cylinders  are 
as  their  altitudes,  when  their  bases  are  equal.  And.  by  the 
corollary  to  the  last  theorem,  they  are  as  their  bases,  when 
their  altitudes  are  equal.  Therefore,  universally,  when  nei- 
ther are  equal,  they  are  to  one  another  as  the  product  of  their 
bases  and  altitudes.  And  hence  also  these  products  are  the 
proper  numeral  measures  of  their  quantities  or  magnitudes. 


THEOREM  CXI. 

Similar   Prisms   and   Cylinders   are   to    each    other,    as    the 
Cubes  of  their  Altitudes,  or  of  any  other  Like  Linear  Di- 
mensions. 
Let  abcd,    efgh,  be   two   similar  prisms  ;    then  will  the^ 

prism  CD  be  to  the  prism  gh,  as  ab^  to  ef*  or  ad^  to  eh  3. 

For 


THEOREMS. 


347 


O 


For  the  solids  are  to  each  other  as 
the  product  of  their  bases  and   alti-  ^  . 
tudes    (th.    1 10,   cor.   2),   that  is,  as     N 
AC  .  AD  to  EG  .  EH.     But  the   bases,  H^''^'*^ 

being   similar    planes,  are    to    each  [S^ / 

other   as   the   squares   of  their    like 
sides,    that   is,    ac   to   eg    as   ab^    to  ^j^ 

ef2  ,  therefore  the  solid  cd  is  to  the        _ 

solid  GH,   as  ab3    .  ad  to  ef^    .  eh.       B  IF 

B^it  ED   and  FH,  being  similar  planes,  have  their  like  sides 

proportional,   that   is,    ab   :  ef   :  :  ad  :   eh, 

or  ab2  :  ef2  :  :  ad^  :  eh^  :  therefore  ab^  :  ad  :  ef^  :  eh  : :  ab  ' :  ef ' , 
or  :  :  ad^  :  eh^  ;,  oonseq.  the  solid  cd  :  solid  gh  :  :  ab^  ; 
|sf3  :  :  ad3  :  eh'.  Q*  e.  d.^ 


^ 


& 


THEOREM  CXIL 


In  any  Pyramid,  a  Section  Parallel  to  the  Base  is  similar  to 
the  Base  ;  aud  these  two  planes  are  to  each  other  as  the 
Squares  of  their  Distances  from  th^  Vertex. 

Let  abcd  be  a  pyramid,  and  efg  a  sec- 
tion parallel  to  the  base  bcd,  also  aih  a 
'line  perpendicular  to  the  two  planes  at  h 
and  I  :  then  will  bd,  eg,  be  two  similar 
planes,  and  the  plane  bd  will  be  to  the 
plane  eg,  as  ah^  to  ai^. 

For,  join  ch,  fi.  Then,  because  a  plane 
cutting  two  parallel  planes,  makes  parallel 
sections  'th.  105),  therefore  the  plane  abc, 
meeting  the  two  parallel  planes  bd,  eg,  makes  the  sections 
bc,  ef,  parallel  :  In  like  manner,  the  plane  acd  makes  the 
sections  cd,  fg,  parallel.  Again,  because  two  pair  of  paral- 
lel lines  make  equal  angles  (th.  104),  the  two  ef,  fg,  which 
are  parallel  to  bc,  cd,  make  the  angle  efg  equal  the  angle 
BCD.  And  in  Hke  manner  it, is  shown,  that  each  angle  in  the 
plane  eg  is  equal  to  each  angle  in  the  plane  bd,  aud  conse- 
quently those  two  planes  are  equiangular. 

Again,  the  three  lines  ab,  ac,  ad,  making  with  the 
parallels  bc,  ef,  and  cd,  fg.  equal  angles  (th.  14),  and 
the  angles  at  a  being  common,  the  two  triangles  abc,  aef, 
are  equiangular,  as  also  the  two  triangles  acd,  afg,  and 
have  therefore  their  like  sides  proportional,  namely,  -  -  - 


348 


GEOMETRY. 


AC  :  AF  :  :  BC  ;  EF  :  :  CD  :  fg.  And  ia 
like  manner  it  may  be  shown,  that  all  the 
lines  in  the  plane  fg,  are  proportional  to  all 
the  corresponding  lines  in  the  base  bd. 
Hence  these  iwo  planes,  having  their  an- 
gles equal,  and  their  sides  proportional,  are 
similar,  by  def.  68.  

But,  similar  planes  being  to  each  other  as  the  squares  of 
their  like  sides,  the  plane  bd  :  eg  :  :  bc^  :  ef^,  or  :  :  ac^: 
AF^,  by  what  is  shown  above.  Also,  the  two  triangles 
AHC,  AiF,  having  the  angles  h  and  i  right  ones  (th.  98), 
and  the  angle  a  common,  are  equiangular,  and  have  there- 
fore their  like  sides  proportional,  namely,  ac  :  af  :  :  ah  :  ai, 
or  Ac3  :  af2  :  :  ah^  :  ai^.  Consequently  the  two  planes 
BD,  EG,  which    are    as    the   former  squares   ac^,   af^,    will 

be   also  as    the    latter  squares    ah^,  ai^,   that  is, 

:bd   :  EG   :  :  ah^    :  ai^.  q,.  e.  d. 


THEOREM  CXIII. 


In  a  Cone,  any  Section  Parallel  to  the  Base  is  a  Circle  ;  and 

this  Section  is  to  the  Base,  as  the  Squares  of  their  Distances 

from  the  Vertex. 

Let  ABCD  be  a  cone,  and  ghi  a  section 
parallel  to  the  base  bcd  ;  then  will  ghi  be 
a  circle,  and  bcd,  ghi,  will  be  to  each  other, 
as  the  squares  of  their  distances  from  the 
vertex. 

For,  draw  alf  perpendicular  to  the  two 
parallel  planes  ;  and  let  the  planes  ace, 
ADE,  pass  through  the  axis  of  the  cone 
ake,  meeting  the  section  in  the- three  points 

H,  I,-K. 

Then,  since  the  section  ghi  is  parallel  to  the  base  bcd,  and 
the  planes  ck,  dk,  meet  them,  hk  is  parallel  to  ce,  and  ik  to 
de  (th  105).  And  because  the  triangles  formed  by  these 
lines  are  equiangular,  kh  :  ec  :  :  ak  :  ae  :  :  ki  :  ed.  But 
EC  is  equal  to  ed,  being  radii  of  the  same  cir<  le  ;  therefore 
Ki  is  also  equal  to  kh.  And  the  same  may  be  shown  of  any 
other  lines  drawn  from  the  point  k  to  the  perimeter  of  the 
■section  ghi,  which  is  therefore  a  circle  (def.  45). 

Again,  by  similar  triangle?,  al  :  af  :  :  ak  :  ae  or 
Ki  :  ED,  hence  al^  :  af^  :  :  ki^  :  fd^  ;  but  ki^  :  ed^ 
circle  ghi  :  circle  bcd  (th.  93)  j  therefore  al^  :  af^ 
circle  ghi  :  circle  bgd.  q    e.  d. 

THEOKEM 


THEOREMS. 


349 


THEOREM  CXIV. 


All  Pyramids,  and  Cones,  of  Equal   Basos  and  Altitudes,  are 
Equal  to  one  another. 


Let  ABC,  DEF,  be 
any  pyramids  and 
cone,  of  equal  bases 
Bc,  EF,  and  equal  al- 
titudCvS  AG,  DH  ;  then 
will  the  pyramids 
and  cone  abc  and 
DEF,  be  equal. 


For,  parallel  to  the 
bases  and  at  equal  distances   an,  do,  from  the  vertices,  sup- 
pose the  planes  ik,  lm,  to  be  drawn. 

Then,     by    the     two     preceding    theorems, - 

Do2  :  dh2  :  :  lm  :  ef,  and 
AN^  :  ag2  :  :  IK   :  bc. 

But  since  an2,  ag^^  are  equal  to  do^,  dh^, 

therefore  ik  :  bc  :  :  lm  :  ek.  But  bc  is  equal  to  ef, 
by  hypothesis  ;  therefore  ik  is  also  equal  to  lm. 

In  like  manner  it  is  shown,  that  any^  other  sections,  at  equal 
distance  from  the  vertex,  are  equal  to'each  other. 

Since  then,  every  section  in  the  cone,  is  equal  to  the  cor- 
responding section  in  the  pyramids,  and  the  heights  are  equal, 
the  solids  abc,  def,  composed  of  all  those  sections,  must  be 
#qual  also.  q.  e.  d. 

THEOREM  CXV. 


Every  Pyramid  is  the  Third  Part  of  a  Prism  of  the  Same  Base 
and  Altitude. 

Let  abcdef  be  a  prism,   and  bdef  a  A 

pyramid,  on  the  same  triangular  base 
DEF  :  then  will  the  pyramid,  bdef  be  a 
third  part  of  the  prism  abcdef. 


E 


/ 


%: 


For,  in  the  planes  of  the  three  sides  of 
the  prism,  draw  the  diagonals  bf,  bd,  cd. 
Then  the  two  planes  bdp,  bcd,  divide  the 
whole  prism  into  the  three  pyramids  bdef,  dabc,  dbcf,  which 
are  proved  to  be  all  equal  to  one  another,  as  follows. 

Since  the  opposite  ends  of  the  prij^m  are  equal  to  each  other, 
the  pyramid  whose  base  is  abc  and  vertex  d,  is  equal  to  the 

pyramitl 


350 


GEOMETRY. 


pyramid  whose  base  is  def  and  vertex  b 
(th.  114),  being  pyramids  of  equal  base  and 
altitude. 

But  the  latter  pyramid,  whose  base  is^ 
DEF  and  vertex  b,  is  the  same  solid  ias  the 
pyramid  whose  base  is  bef  and  vertex  d, 
and  this  is  equal  to  the  third  pyramid 
whose  base  is  b<  f  and  vertex  d,  being  py- 
ramids of  the  same  altitude  and  equal  bases 
bef,  bcf.  ,  . 

Consequently  all  the  three  pyramids,  which  compose  the 
•prism,  are  equal  to  each  other,  and  each  pyramid  is  the  third 
part  of  the  prism,  or  the  prism  is  tripje  of  the  pyramid. 

Q.  E.  rj. 

Hence  also,  every  pyramid,  whatever  its  figure  may  be,  is 
the  third  part  of  a  prism  of  the  same  base  and  altitude  ;  since 
the  base  of  the  prism,  whatever  be  its  figure,  may  be  divided 
into  triangles,  and  the  whole  solid  into  triangular  prisms  and 
pyramids. 

Corol.  Any  cone  is  the  third  part  of  a  cylinder,  or  of  a 
prism,  of  equal  base  and  altitude  ;  since  it  has  been  proved 
that  a  cylinder  is  equal  to  a  prism,  and  a  cone  equal  to  a  pyra- 
mid, of  equal  base  and  altitude. 

Scholium.  Whatever  has  been  demonstrated  of  the  propor- 
tionality of  prisms,  or  cylindere,  holds  equally  true  of  pyra- 
mids, or  cones  ;  the  former  bemg  always  triple  the  latter  ;  viz. 
that  similar  pyramids  or  cones  are  as  the  cubes  of  their  like 
linear  sides,  or  diameters,  or  ailtiludes,  &,c.  And  the  same 
for  all  similar  solids,  whatever,  viz.  that  they  are  in  proportion^ 
to  each  other,  as  the  cubes  of  their  like  linear  dimensions, 
since  they  are  composed  of  pyramids  every  way  similar. 


THEOREM  CXVI. 


If  a  Sphere  be  cut  by  a  Plane,  the  Section  will  be  a  Circle. 

Let  the  sphere  aebf  be  cut  by  the 
plane  adb  ;  then  will  the  section  adb  be 
a  circle. 

Draw  the  chord  ab,  or  diameter  of 
the  section  ;  perpendicular  to  which,  or 
to  the  section  adb,  draw  the  axis  of  the 
sphere  ecgf,  through  the  centre  c, 
which  will  bisect  the  chord  ab  in  the 
point  G   (th.  41).     Also,  join  ca,  cb  ; 

and 


THEOREMS. 


351 


and  draw  cd,  «d,  to  any  point  d  in  the  perimeter  of  the  sec- 
tion AOB. 

Then,  because  cg  is  perpendicular  to  the  plane  a»b,  it  is 
perpendicular  both  to  ga  and  gd  (def.  90).  So  that  cga,  cgd, 
are  two  risfht-angled  triangles,  having  the  perpendicular  cc 
cornmon.  and  the  two  hypothenuses  ca,  cd,  equal,  being  both 
radii  of  the  sphere  ;  therefore  the  third  sides  ga,  gd,  are  also 
equal  (cor.  2,  th.  34).  In  like  manner  it  is  shown,  that  any 
other  line,  drawn  from  the  centre  g  to  the  circumference  of 
the  section  adb,  is  equal  to  ga  or  gb  ;  consequently  that  sec- 
tion is  a  circle. 

Corol.  The  section  through  the  centre,  is  a  circle  having 
the  same  centre  and  diameter  as  the  sphere,  and  is  called  a 
great  circle  of  the  sphere  ;  the  other  plane  sections  being  \iU 
tie  circles. 


THEOREM  CXVII. 


Every  Sphere  is  Two-Thirds  of  its  Circumscribing  Cylinder 


^IC 

"^ 

/^\ 

^.^ 

K 

y 

B 


Let  abcd  be  a  cylinder,  circumscrib-  a 

ing  the  sphere   efgh  ;    then   will   the 
sphere  efgh  be  two-thirds  of  the  cylin-         0 
der  ABt.D. 

For,  let  the  plane  ac  be  a  section  of 
the   sphere   and  cylinder  through   the 

centre  i.     Join  ai,  bi.     Also,  let  fih  be  ^ 

parallel  to  ad  or  bc,  and  eio  and  kl  pa-  D        H 

rallel  to  ab  or  dc,  the  base  of  the  cy- 
linder ;  the  latter  line  kl  meeting  bi  in  m,  and  the  circular 
section  of  the  sphere  in  n. 

Then,  if  the  whole  plane  hfbc  be  conceived  to  reVolve 
about  the  line  hf  as  an  axis,  the  square  fg  will  describe  a 
cylinder  ag,  and  the  quadrant  ifg  will  describe  a  hemisphere 
EFG,  and  the  triangle  ifb  will  describe  a  cone  iab.  Also,  in 
the  rotation,  the  three  lines  or  parts  kl,  kn,  km,  as  radii,  will 
describe  corresponding  circular  sections  of  those  solids,  name  • 
ly,  KL  a  section  of  the  cylinder,  KJf  a  section  of  the  sphere, 
and  KM  a  section  of  the  cone. 

Now,  FB  being  equal  to  fi  or  ig,  and  kl  parallel  to  fb, 
then  by  similar  triangles  ik  is  equal  to  km  (th.  82).  And  since, 
in  the  right-anglv-d  triangle  ikn,  in^  is  equal  to  ik^  -f.  kx^ 
(th.  34)  ;  and  because  kl  is  equal  to  the  radius  le  or  in,  and 


352 


GEOMETRY. 


A 
O 


XQ- 


D 


KM  =  IK,  therefore  ki.^  is  equal  to  km^ 
-f  KN^,  or  the  square  of  the  longest  ra- 
dius, of  the  said  circular  sections,  is 
equal  to  the  sum  of  the  squares  ol  the 
two  others.  And  because  circles  are  to 
each  other  as  the  squares  of  their  diam- 
eters, or  of  their  radii,  therefore  the 
circle  described  by  kl  is  equal  to  both 
the  circles  des<  iliied  by  km  and  kn  ;  or 
the  section  of  the  cylinder,  is  equal  to  both  the  corresponding 
sections  of  the  sphere  and  cone.  And  as  this  is  always  the 
case  in  every  parallel  position  of  kl,  it  follows,  that  the  cylin- 
der EB,  which  is  composted  of  all  the  former  s^ections,  is  equal 
to  the  hemisphere  efg  and  cone  iab,  which  are  composed  of 
of  all  the  latter  sections. 

But  the  cone  iab  is  a  third  part  of  the  cylinder  fb  (cor.  2, 
th.  116)  ;  consequentl)  the  heoiisphere  efg  is  equal  to  the 
remaining  two-thirds  ;  or  the  whole  sphere  efgh  equal  to  two- 
thirds  ot  the  whole  cylinder  abcd.  q.  e.  d 

Corol.  1.  A  cone,  hemisphere,  and  cylinder  of  the  same 
base  and  altitude,  are  to  each  other  as  the  numbers  1,  2,  3. 

Corol.  2.  All  spheres  are  to  each  other  as  the  cubes  of  their 
diameters  ;  all  these  being  like  parts  of  their  circumscribing 
cylinders. 

Corol.  3.  From  the  foregoing  demonstration  it  also  appears^ 
that  the  spherical  zone  or  frustrum  egnp,  is  equal  to  the  diffe- 
rence between  the  cylinder  eglo  and  the  cone  imq.  all  of  the 
same  common  height  ik.  And  that  the  spherical  segment  pfn, 
is  equal  to  the  difference  between  the  cylinder  ablo  and  the 
conic  frustrum  a^imb,  all  of  the  same  common  altitude  fk. 


PROBLEMS. 


[353} 

PR0BLJ2M&. 
PROBLEM  I. 


To  Bisect  a  Line  ab 


that  is,  to  divide  it  iato  two  Equal 
Parts. 


From  the  two  centres  a  and  b,  with 
any  equal  radii,  describe  arcs  of  circles, 
intersecting  each  other  in  c  and  d  ;  and 
draw  the  line  cd,  which  will  bisect  the 
given  line  ab  in  the  point  e. 

For,  draw  the  radii  ao,  bc,  ad,  bd. 
Then,  because  all  these  four  radii  are 
equal,  and  the  side  cd  common,  the  two 
triangles  acd,  bcd,  are  mutually  equilateral  :  consequently 
they  are  also  mutually  equiangular  (th.  5),  and  have  the  angle 
ACE  equal  to  the  angle  bce. 

Hence,  the  two  triangles  ace,  bce,  having  the  two  sides 
AC,  CE,  equal  to  the  two  sides  bc,  ce,  and  their  contained  an- 
gles equal,  are  identical  (th.  1),  and  therefore  have  the  side 
ae  equal  to  eb.  q.  e.  d. 


PROBLEM  II. 
To  Bisect  an  Angle  bac. 

From  the  centre  a,  with  any  radius,  de- 
scribe an  arc,  cutting  oflf  the  equal  lines 
ad,  ae  ;  and  from  the  two  centres  d,  e, 
with  the  same  radius,  describe  arcs  inter- 
secting in  F  ;  then  draw  af,  which  will  bi- 
sect the  angle  a  as  required. 

For,  join  df,  ef.  Then  the  two  tri- 
angles adf,  aef,  having  the  two  sides 
AD,  DF,  equal  to  the  two  ae,  ef  (being  equal  radii),  and  the 
side  AF  common,  they  are  mutually  equilateral ;  consequently 
they  are  also  mutually  equiangular  (th.  3),  and  have  the  angle 
BAF  equal  to  the  angle  caf. 

ScholiuiYi.     In  the  sacne  manner  is  an  arc  of  a  circle,  bi- 
sected. 

Vot,  I.-  4« 

PROBt^M 


354 


GEOMETHY. 


PROBLEM  in. 


EB 
Then  the  two 


At  a  Given  Point  c,  in  a  Line  ab,  to  Erect  a  Perpendicular. 

From  the  given  point  c,  with  any  radius, 
cut  off  any  equal  parts  cd,  ce,  of  the  given 
line  ;  and,  from  the  two  centres  d  and  e, 
with  any  one  radius,  describe  arcs  intersect- 
ing in  F  ;  then  join  cf,  which  will  be  per- 
pendicular as  required. 

Tor,  draw  the  two  equal  radii  df,  ef. 
triangles  CDF,  cef,  having  the  two  sides  cd,  df,  equal  to 
the  two  CE,  EF,  and  cf  common,  are  mutually  equilateral  ; 
consequently  they  are  also  mutually  equiangular  (th.  5),  and 
have  the  two  adjacent  angles  at  c  equal  to  each  other  ;  there- 
fore the  line  cf  is  perpendicular  to  ab  (def.  11). 

Otherwise. 
When  the  Given  Point  c  is  near  the  End  of  the  line. 

From  any  point  d,  assumed  above  the 
line,  as  a  centre,  through  the  given  point 
c  describe  a  circle,  cutting  the  given  Hne 
at  E  ;  and  through  e  and  the  centre  d, 
draw  the  diameter  edf  ;  then  join  cf, 
which  will  be  the  perpendicular  required. 

For  the  angle  at  c,  being  an  angle  in  a  semicircle,  is  a 
right  angle,  and  therefore  the  line  cf  is  a  perpendicular 
(by  def  15). 

PROBLEM  lY. 

From  a  Given  point  a  to  let  fall  a  Perpendicular  on  a  given 
Line  bc. 

From  the  given  point  a  as  a  centre,  with 
any  convenient  radius,  describe  an  arc,  cut- 
ting the  given  line  at  the  two  points  d  and 
£  ;  and  from  the  two  centres  d,  e,  with 
any  radius,  describe  two  arcs,  intersecting 
at  f  ;  then  draw  agf,  which  will  be  perpen- 
dicular to  bc  as  required- 

For,  draw   the    equal    radii   ad,   ae,   and 
DF,  EF.     Then  the  two  triangles  adf,  aef,  having  the  two 
sides  ad,  df,  equal  to  the  two  ae,  ef,  and  af  common,  are 

mutually 


PROBLEMS.  356 

mutually  equilateral  ;  consequently  they  are  also  mutually 
equiangular  (th.  5),  and  have  the  ani^le  dag  equal  the  angle 
EAG.  Hence  then,  the  two  triangles  adg,  aeg,  having  the 
two  sides  ad,  ag,  equal  to  the  two  ae,  ag,  ahd  their  included 
angles  equal,  are  therefore  equiangular  (th  1),  and  have  the 
angles  at  g  equal  ;  consequently  ag  is  perpendicular  to  bc 
(def.  11). 

Otherwise. 

When  the  Given  Point  is  nearly  Opposite  the   end  of  the 
Line. 

From  any  point  d,  in  the  given  line 
Be,  as  a  centre,  descrihe  the  arc  of  a 
circle  through  the  given  point  a,  cutting 
BC  in  B  ;  and  from  the  centre  e,  with  the 
radius  ea,  describe  another  arc,  cutting 
the  former  in  f  ;  then  draw  agf,  which 
will  be  perpendicular  to  bc  as  required.  F 

For,  draw  the  equal  radii  da,  df,  and  ea,  ef.  Then  the 
two  triangles  dae,  dfe,  will  be  mutually  equilateral  ;  conse- 
quently they  are  also  mutually  equiangular  (th.  5),  and  have 
the  angles  at  d  equal.  Hence,  the  two  triangles  dag,  dfg, 
having  the  two  sides  da,  dg,  equal  to  the  two  df,  dg,  and  the 
included  angles  at  d  equal,  have  also  the  angles  at  g  equal 
(th.  1)  ;  consequently  those  angles  at  g  are  right  angles,  and. 
the  line  ag  is  perpendicular  to  dg. 


PROBLEM  V^ 

At  a  Given  Point  a,  in  a  Line  ab,  to  make  an  Angle  Equal 
to  a  Given  Angle  c. 

From  the  centres  a  and  c,  with  any  one  JEi 

radius,    describe  the  arcs  de,  fg.     Then, 
with  radius  de,  and  centre  f,  describe   an 


arc,  cutting  fg    in  g.       Through    g   draw  C  D 

the  line  ag,  and  it  will  form  the  angle  re-  ^^ 

quired.  '  ^,^^  \ 

For,  conceive  the   equal  lines  or  radii,         A  FB 

de,  fg,  to  be  drawn.  Then  the  two  triangles  cde,  afg,  bemo" 
mutually  equilateral,  are  mutually  equiangular  (th.  5),  and 
have  the  angle  at  a  equal  to  the  angle  c. 

PROBLEM 


V 


366  '  GEOMETRY. 

PROBLEM  VI. 

Through  a  Given  Point  a,  to  draw  a  Line  Parallel  to  a  Given 
Line  bc. 

From  the  given  point  a  draw  a  line  ad  EA 

to  any  point  in  the  given  line  bc  Then 
draw  the  hne  eaf  making  the  angle  at  a 
equal  to  the  angle  at  d  (by  prob.  5)  ;  so  jg~"         jS^ 

shall  EF  be  parallel  to  bc  as  required. 

For.  the  angle  d  being  equal  to  the  alternate  angle  a,  th^ 
lines  BC,  EF,  are  parallel,  by  th.  13. 

PROBLEM  VII. 

To   Divide   a  Line  ab  into  any  proposed  Number  of  Equal 
Parts. 


m 


Draw  any  other  line  ac,  forming  any 
iangle  with  the  given  line  ab  ;  on  which 
set  off  as  many  of  any  equal  parts,  ad,  de, 
EF,  Fc,  as  the  line  ab  is  to  be  divided  into. 
Join  bc  ;  parallel  to  which  draw  the  other  A.  I  H  &  B 

lines  FG,  EH,  Di  :    then  these  will  divide 
ab  in  the  manner  as  required. — For  those  parallel  lines  divide 
l)oth  the  sides  ab,  ac,  proportionally,  by  th.  82. 

PROBLEM  \1IL 

To  find  a  Third  Proportional  to  Two  given  Lines  ab,  ac,v 

PtACE  the  two  given  lines  ab,  ac, 

forming  any  angle  at  a  ;  and  in  ab  take  A — B 

?ilso  AD   equal  to   ac     Join  bc,  and  A                -C 

draw  DE  parallel  to  it ;  so  will  ae  be  -.  C 

the  third  proportional  sought.            .  ""^^ 

For,  because  of  the  parallels  bc,  de, 


the  two  lines  ab,  ac,  are  cut  propor-  ^  " 

tionally    (th.   82)  ;  90  that  ab  :  ac   :  :  ad  or  ac  :  ae  ;  there-- 
fbre  AE  is  the  third  proportional  to  ab,  ac. 

PROBLEM  IXt 

To  find  a  Fourth  Proportional  to  three  Lines  ab,  ac,  ad. 

Place  two  of   the  given  lines  ab,  ac,  making  any  angle 
at  A  ;  also  place  ad  on  ab.     Join  bc  j  and  parallel  to  it  draw 

de  : 


PROBLEMS. 


367 


Dfi :  SO  shall  AE  be  the  fourth  propor- 
tional as  required. 

For,  because  of  the  parallels  bc,  de, 
the  two  sides  ab,   ac,  are  cut  propor- 
tionally (th.  82)  ;  so  that     . 
AB  :  AC  :  :  AD  :  ae. 


A — 

A 

B 

A 

D^ 

^r\ 

A. 

"      I>      B 

PROBLEM  X. 

To  find  a  Mean  Proportional  between  Two  Lines  ab,  bc. 

Place  ab,  bc,  joined  in   one  straight         p^ g 

line  AC  :  on  which  as  a  diameter,  des- 
cribe the  semicircle  adc  ;  to  meet  which 
erect  the  perpendicular  bd  ;  and  it  will 
be  the  mean  proportional  sought,  be- 
tween AB  and  Bc  (by  cor.  th.  87). 


D— C 


OB    0 


PROBLEM  XI. 
To  find  the  Centre  of  a  Circle. 


Draw  any  chord  ab  ;  and  bisect  it  per- 
pendicularly with  the  line  cd,  which  will 
be  a  diameter  (th.  41,  cor.).  Therefore, 
&D  bisected  into  o,  will  give  the  centre, 
as  required. 


PROfiLEA^  XII. 

To  describe  the  Circamference  of  a  Circle  through  Three 
Given  Points  a,  b,  c. 

From  the  paiddle  point  b  draw  chords 
BA,  BC  to  the  two  other  points,  and  bi- 
sect these  chords  perpendicularly  by 
lines  meeting  in  o,  which  will  be  the 
centre.  Then  from  the  centre  o,  at  the 
distance  of  any  one  of  the  points,  as  oa, 
describe  a  circle,  and  it  will  pass  through 
the  two  other  points  b,  g,  as  required. 

For,  the  two  right-angled  triangles  oad,  obd,  havino- the 
sidfis  AD,  DB,  equal  (by  constr.),  and  od  common  with  the 
included  right  angles  at  d  equal,  have  their  third  sides  oa, 
OB,  also  equal  (th.  J).  And  in  like  manner  it  is  shown,  that 
oc,  is  equal  to  ob  or  oa.  So  that  all  the  three  oa,  ob,  oc, 
^eing  equal,  will  be  radii  of  the  same  circle. 

PROBLEM 


358 


GEOMETRY. 


PROBLEM  XIII. 

To  draw  a  Tangent  to  a  Circle,  through  a  Giren  Point  i. 

When  the  given  point  a  is  in  the  cir- 
cumference of  the  circle  :  Join  a  and  the 
centre  o  ;  perpendicular  to  which  draw 
BAG,  and  it  will  be  the  tangent,  by  th.  46. 

But  when  the  given  point  a  is  out  of 
the  circle  :  draw  ao  to  the  centre  o  ; 
on  which  as  a  diameter  describe  a  semi- 
circle, cutting  the  given  circumference 
in  D  ;  through  which  draw  badc,  which 
will  be  the  tangent  as  required. 

For,  join  do.  Then  the  angle  ado, 
in  a  semicircle,  is  a  right  angle,  and 
consequently  ad  is  perpendicular  to  the 
radius,  do,  or  is  a  tangent  to  the  circle 
(th.  4tJ) 

PROBLEM  XIV.  • 

On  a  Given  Line  b  to  describe   a  Segment  of  a  Circle,  to 
Contain  a  Given  Angle  c. 

At  the  ends  of  the  given  line  make 
angles  DAB,  dba,  each  equal  to  the 
given  angle  c.  Then  draw  ae,  be, 
perpendicular  to  ad,  bd  ;  and  with  the 
centre  e  and  radius  ea  or  eb,  describe 
a  circle  ;  so  shall  Afb  be  the  segment 
required,  as  an  angle  f  made  in  it  will 
be  equal  to  the  given  angle  c. 

For,  the  two  lines  ad,  bd,  being  per- 
pendicular to  the  radii  ea,  eb  (by  constr.),  are  tangents  to  the 
circle  (th.  46)  ;  and  the  angle  a  or  b,  which  is  equal  to  the 
given  angle  c  by  construction,  is  equal  to  the  angle  f  in  the 
alternate  segment  afb  (th.  63). 


PROBLEM  XV. 


To  Cat  ofl'  a  Segment  from  a  Circle,  that  shall  Contsan  a 
Given  Angle  c. 

Draw  any  tangent  ab  to  the  given 
circle  ;  and  a  chord  ad  to  make  the 
angle  dab  equal  to  the  given  angle  c  ; 
then  DEA  will  be  the  segment  required, 
an  angle  e  made  in  it  being  equal  to 
the  given  angle  e. 


PROBLEMS. 


35» 


For  tbe  angle  a,  made  by  the  tans;ent  and  chord,  which  is 
equal  to  the  given  angle  c  by  construction,  is  also  equal  to  any 
ti^e  E  in  the  alternate  segment  (th.  53). 


PROBLEM  XVI. 
To  make  an  Equilateral  Triangle  on  a  Giren  Line  ab. 

From  the  centres  a  and  b,  with  the 
distance  ab,  describe  arcs,  intersecting 
in  c.  Draw  ac,  bc,  and  abc  will  be  the 
equilateral  triangle 

For  the  equal  radii  ac,  bc,  are,  each 
of  them,  equal  to  ab. 


PROBLEM  XVn. 


To  make  a  Triangle  with  Three  Given  Lines  ab,  ac,  bc. 


With  the  centre  a,  and  distance  ac, 
describe  an  arc.  With  the  centre  b, 
and  distance  bc,  describe  another  arc, 
cutti;)}i;  the  former  in  c  Draw  ac,  bc, 
and  ABC  will  be  the  triangle  required. 

For  the  radii,  or  sides  of  the  triangle, 
AC,  BC,  are  equal  to  the  given  lines  ac, 
BC,  by  construction. 


PROBLEM  XVIIL 


To  make  a  Square  on  a  Given  Line  ab. 

Raise  ad,  bc,  each  perpendicular  and 
equal  to  ab  ;  and  join  dc  ;  s'o  Shall  abcd 
be  the  square  sought. 

For  all  the  three  sides  ab,  ad,  bc,  are 
equal,  by  the  construction,  and  dc  is  equal 
an«)  parallel  to  ab  (by  th.  24)  ;  so  that  all  j 

the  four  sides  are  equal,  and  the  opposite 
ones  are  parallel.  Again,  the  angle  a  or  b,  of  the  parallelo- 
gram, being  a  right  angle,  the  angles  are  all  right  ones  (cor. 
1,  th.  22).  Henre,  then,  the  figure,  having  all  its  sides  equal, 
sind  all  its  angles  right,  is  a  square  (def.  34). 


B 


PROBLEM 


36d 


GEOMETRY. 


PROBLEM  XrX. 

To  make  a  Rectangle,  or  a  Parallelogram,  of  a  Given  Lengtk 
and  Breadth,  ab,  bc. 

Erect  ad,  bc,  perpendicular  to  ab,  and  I>  C 

each   equal   to  bc  ;  then  join  dc,  and  it  is 
done. 

The  demonstration  is  the  same  as  the  last  j 

prohlem.  B- 

And  in  the  same  manner  is  descrihed  any  oblique  parallel- 
ogram, only  drawing  ad  and  bc  to  make  the  given  oblique  an 
gle  with  ab,  instead  of  perpendicular  to  it. 


B 


PROBLExM  XX. 

To  Inscribe  a  Circle  in  a  Given  Triangle  abc. 
Bisect  any  two  angles  a  and  b,  with 
the  two  lines  ad,  bd.  From  the  inter- 
section D,  which  will  be  the  centre  of 
the  circle,  draw  the  perpendiculars  de, 
DF,  dg,  and  they  will  be  the  radii  of  the 
circle  required. 

For,  since  the  angle   dae  is  equal   to 
the   angle  dag,  and  the  angles  at  e,  g, 

right  angles  (by  constr.),  the  two  triangles  ade,  adg,  are  equi- 
angular ;  and,  having  also  the  side  ad  common,  they  are  iden- 
tical, and  have  the  sides  de,  dg,  equal  (th.  2).  In  like  man- 
ner it  is  shown,  that  df  is  equal  to  de  or  dg. 

Therefore,  if  with  the  centre  d,  and  distance  de,  a  circle 
be  described,  it  will  pass  through  all  the  three  points  e,  f,  g, 
in  which  points  also  it  will  toucuj  the  three  sides  of  the  triangle 
(th.  46),  because  the  radii  de,  df,  dg,  are  perpendicular  t© 
them. 


PROBLEM  XXL 

To  Describe  a  Circle  about  a  Given  Triangle  abc 

Bisect  any  two  sides  with  two  of  the 
perpendiculars  de,  df,  dg,  and  d  will  be 
the  centre. 

For,  join  DA,  DB,  dc.  Then  the  two 
right-angled  triangles  dae,  dbe,  have  the 
two  sides  de,  ea,  equal  to  the  two  de, 
eb,  and  the  included  angles  at  e  equal  : 
those  two  triangles  are  therefore  identical 


(th. 


PROBLEMS. 


361 


(th.  1),  and  have  the  side  da  equal  to  db.  In  like  manner 
it  is  shown,  thaf  dc  is  also  equal  to  da  or  db.  So  that  all 
the  three  da,  db,  dc,  being  equal,  they  are  radii  of  a  circle 
passing  through  a,  b,  and  c. 

PROBLEM  XXn. 


To  Inscribe  an  Equilateral  Triangle  in  a  Given  Circle- 

Through  the  centre  c  draw  any  oTa- 
meter  ab.  From  the  point  b  as  a  centr<|| 
with  the  radius  bc  of  the  given  circle, 
describe  an  arc  dce.  Join  ad,  ae,  de, 
and  ADE  is  the  equilateral  triangle  sought. 

For,  join  db,  do,  eb,  eg.  Then  dcb 
is  an  equilateral  triangle,  having  each 
side  equal  to  the  radius  of  the  given  circle. 
In  like  manner,  bce  is  an  equilateral  triangle.  But  the  angle 
ADE  is  equal  to  the  angle  abe  or  cbe,  standing  on  the  same 
arc  AE  ;  also  the  angle  aed  is  equal  to  the  angle  cbd,  on,  the 
same  arc  ad  ;  hence  the  triangle  dae  has  two  of  its  angles, 
ADE,  aed,  equal  to  the  angles  of  an  equilateral  triangle,  and 
ther^ore  the  third  angle  at  a  is  also  equal  to  the  same  ;  so 
that  triangle  is  equiangular,  and  therefore  equilateral. 


PROBLEM  xxnr. 


To  Inscribe  a  Square  in  a  Given  Circle. 

Deaw  two  diameters  ac,  bd,  crossing 
at  right  angles  in  the  centre  ».  Then 
join  the  four  extremities  a,  b,  c,  d,  with 
right  lines,  and  these  will  form  the  in- 
scribed square  abcd. 

For,  the  four  right-angled  triangles 
aeb,  beg,  ged,  »ea,  are  identical,  be- 
cause they  have  the  sides  ea,  eb,  eg,  ed, 
all  equal,  being  radii  of  the  circle,  and 
the  four  included  angles  at  e  all  equal, 
being  right  angles,  by  the  construction.  Therefore  all  their 
third  sides  ab,  bg,  cd,  da,  are  equal  to  one  another,  and  the 
figure  ABCD  is  equilateral.  Also,  all  its  four  angles,  a,  b,  c, 
D,  are  right  ones,  being  angles  in  a  semicircle.  Consequently 
the  figure  is  a  square. 


VOB.  h 


47 


PROBLPM 


f 

^ 

\^ 

V 

362  GEOMETRY. 

PROBLEM  XXIV. 
To  Describe  a  Square  about  a  Given  Circle. 

Draw  two  diameters  ac,  bd,  crossing 
at  right  angles   in   the  centre  e.     Then        ^     J^y^^   ? 
through  their  four  extremities  draw  fg, 
iH,  parallel  to  ac,  and  fi,  gh,  parallel  to 
BD,  and  they  will  form  the  square  fghi. 

For,  the  opposite  sides  of  parallelo- 
grams being  equal,  fg  and  ih  are  each  r — ^^  j^  g." 
equal  to  the  diam||ter  ac,  and  fi  and  gh 
each  equal  to  the  diameter  bd  ;  so  that 
the  figure  is  equilateral.  Again,  because  the  opposite  angles 
of  parallelograms  are  equal,  all  the  four  angles  f,  g,  h,  i,  are 
right  angles,  being  equal  to  the  opposite  angles  at  e.  So  that 
the  figure  fghi,  having  its  sides  equal,  and  its  angles  right 
ones,  is  a  square,  and  its  sides  touch  the  circle  at  the  four 
points  a,  b,  c,  d,  being  perpendicular  to  the  radii  drawn  to 
those,  points. 

I>ROBLEM  XXV. 
To  Inscribe  a  Circle  in  a  Given  Square. 

Bisect  the  two  sides  fg,  fi,  in  the  points  a  and  b  (last fig.). 
Then  through  these  two  points  draw  ac  parallel  to  fg  or  ih, 
and  BD  parallel  to  fi  or  gh.  Then  the  point  of  intersection 
E  will  be  the  centre,  and  the  four  lines  ea,  eb,  ec,  ed,  radii 
of  the  inscribed  circle. 

For,  because  the  four  parallelograms  eb*,  eg,  eh,  ei,  have 
their  opposite  sides  and  angles  equal,  therefore  all  the  four 
lines  ea,  eb,  ec,  ed,  are  equal,  being  each  equal  to  half  a 
side  of  the  square.  So  that  a  circle  described  from  the  centre 
E,  with  the  distance  ea,  will  pass  through  all  the  points  a,  b, 
c,  d,  and  will  be  inscribed  in  the  square,  or  will  touch  its  four 
sides  in  those  points,  because  the  angles  there  are  right  ones. 

PROBLEM  XXVL 

To  Describe  a  Circle  about  a  Given  Square, 
(see  fig.  Prob.  xxiii.) 

Draw  the  diagonals  ac,  bd,  and  their  intersection  e  will  be 
the  centre. 

For  the  diagonals  of  a  square  bisect  each  other  (th.  40), 
making  ea,  eb,  ec,  ed,  all  equal,  and  consequently  these 
are  radii  of  a  circle  passing  through  the  four  points  a,  b,  c,  d. 

PROBLEM 


PROBLEMS. 


36d 


Then,  adf 


PROBLEM  XXVII. 
To  Gut  a  Given  Line  in  Extreme  and  Mean  Ratio. 

Let  ab  be  the  giren  line  to  be  divided 

in  extreme  and  mean  ratio,  that  is,  so  as 

that  the  whole  line  may  be  to  the  greater 

part,  as  the  greater  part  is  to  the  less  part. 
Draw  Bc  perpendicular  to  ab,  and  equal 
to  half  AB.  Join  ac  ;  and  with  centre  c 
and  distance  ce,  describe  the  circle  bd  ; 
then  with  centre  a  and  distance  ad,  de- 
scribe the  arc  de  ;  so  shall  ab  be  divided 
in  E  in  extreme  and  mean  ratio,  or  so  that 
AB   :   AE,;  :   ae   ;   eb. 

For,  produce  ac  to  the  circumference  at 
being  a  secant,  and  ab  a  tangent,  because  b  is  a  right  angle  ; 
therefore  the  rectangle  af  .  ad  is  equal  to  ab^  (cor.  1,  th.  61)  ; 
consequently  the  means  and  extremes  of  these  are  proportion- 
al (th.  77),  viz.  ab  :  af  or  ad  -f-  df  :  :  ad  :  ab.  But  ae  is 
equal  to  ad  by  construction,  and  ab  =  2bc  =  df  ;  therefore, 
ab  :  ae  -I-  ab  :  :  ae  :  ab  ;  and  by  division, 
AB   :.  AE   :  :  ak   :  eb. 

PROBLEM  XXVIIL 

To  Inscribe  an  Isosceles  Triangle  in  a  Given  Circle,  that  shall 
have  each  of  the  Angles  at  the  Base  Double  the  Angle  at 
the  Vertex. 

Draw  any  diameter  ab  of  the  given 
circle  ;  and  divide  the  radius  cb,  in  the 
point  d,  in  extreme  and  mean  ratio,  by 
the  last  problem.  From  the  point  b 
apply  the  chords  be,  bf,  each  equal  to 
the  greater  part  cd.  Then  join  ae,  af, 
EF  ;  and  aef  will  be  the  triangle  requir- 
ed. 

For,  the  chords  be,  bf,  being  equal,  their  arcs  are  equal  ; 
therefore  the  supplemental  arcs  and  chords  ae,  af,  are  also 
equal ;  consequently  the  triangle  aef  is  isosceles,  and  has 
the  angle  e  equal  to  the  angle  f  ;  also  the  angles  at  g  are 
right  angles. 

Draw  OF  and  df.  Then,  bc  :  cd  :  :  cd  :  bd,  or 
EC  :  bf  :  :  bf  :  bd  by  constr.  And  ba  :' bf  :  :  bf  :  bg 
(by  tb.  87).  But  bc  =  4^ba  ;  therefore  bg  =  Ibd  =  gd  ; 
therefore  the  two  triangles  gbf,  gdf,  are  identical  (th.  1), 

and 


364 


€frEOMETRY. 


and  each  equiangular  to  abf  and  agf  (th.  87).  Therefofe 
their  doubles,  bfd,  afe,  are  isosceles  and  equiangular,  as 
well  as  the  triangle  bcf  ;  having  the  two  sides  bc,  cf,  equal, 
and  the  angle  b  common  with  the  triangle  bfd.  But  cb 
is  =  DF  or  BF  ;  therefore  the  angle  c  =  the  angle  dfc  (th. 
4)  ;  consequently  the  angle  bdp,  which  is  equal  to  the  sum  of 
these  two  equal  angles  (th.  16),  is  double  of  one  of  them  c  ;  or 
the  equal  angle  b  or  cfb  double  the  angle  c.  So  that  cbf  ifi 
an  isosceles  triangle,  having  each  of  its  two  equal  angles 
double  of  the  third  angle  c.  Consequently  the  triangle  aef 
(which  it  has  been  shown  is  equiangular  to  the  triangle  cbf) 
has  also  each  of  its  angles  at  the  base  double  the  angle  a  at 
the  vertex. 

PROBLEM  XXIX. 

To  Inscribe  a  Regular  Pentagon  in  a  Given  Circle. 

Inscribe  the  isosceles  triangle  abc 
having  each  of  the  angles  abc,  acb, 
double  the  angle  bac  (prob.  28).  Then 
bisect  the  two  arcs  adb,  afc,  in  the 
points  D,  E  ;  and  draw  the  chords  ad, 
»b,  ae,  EC,  so  shall  adbce  be  the  in- 
scribed equilateral  pentagon  required. 

For,  beeause  equal  angles  stand  on 
equal  arcs,  and  double  angles  on  double  arcs,  also  the  angles 
ABC,  ACB,  being  each  double  the  angle  bac,  therefore  the  arcs 
ADB,  AEc,  subtending  the  two  former  angles,  each  one  double 
the  arcs  bc  subtending  the  latter.  And  since  the  two  former 
arcs  are  bisected  in  d  and  e,  it  follows  that  all  the  five  arcs 
AD,  db,  bc,  ce,  ea,  are  equal  to  each  other,  and  consequently 
the  chords  also  which  subtend  them,  or  the  five  sides  of  the 
pentagon,  are  all  equal. 

Note   In  the  construction,  the  points  d  and  e  are  most  easily 
found,  by  applying  bd  and  ce  each  equal  to  bc. 

PROBLEM  XXX. 

To  Inscribe  a  Regular  Hexagon  in  a  Circle. 

Apply  the  radius  ao  of  the  given  circle 
as  a  chord,  ab,  bc,  cd,  &c.  quite  round  the 
circumference,  and  it  will  complete  the 
regular  hexagon  abcdef. 

For,  draw  the  radii  ao,  bo,  co,  do,  eo, 
yo,  completing  six  equal  triangles  ;  of 
which  any  one,  as  abo,  being  equilateral 


PROBLEMS.  365 

by  constr.)it8  three  angles  are  all  equal  (cor.  2,  th.  3),  and 
any  one  of  them,  as  aub,  is  one-third  of  the  whole,  or  of  two 
right  angles  (th.  17),  or  one-sixth  of  four  right  angles.  But 
the  whole  circumference  is  the  measure  of  four  right  angles 
(cor  4,  th.  6).  Therefore  the  arc  ab  is  one-sixth  of  the  cir- 
cumference of  the  circle,  and  consequently  its  chord  ab  one 
aide  of  an  equilateral  hexagon  inscribed  in  the  circle.  And 
the  same  of  the  other  chords. 

Corol,  The  side  of  a  regular  hexagon  is  equal  to  the  radius 
of  the  circumscribing  circle,  or  to  the  chord  of  one-sixth 
part  of  the  circumference. 

PROBLEM  XXXI. 

To  describe  a  Regular  Pentagon  or  Hexagon  about  a  Circle. 

In  the  given  circle  inscribe  a  regular  A^ 

polygon  of  the  same  name  or  number  ~ 

of  sides,  as  abode,  by  one  of  the  forego- 
ing problems.  Then  through  all  its  an- 
gular points  draw  tangents  (by  prob.  13) 
and  these  will  form  the  circumscribing 
polygon  required. 

For,  all  the  chords,  or  sides  of  the 
inscribing  figure,  ab,  bc,  &c.  being  equal,  and  all  the  radii 
OA,  oB,  &c.  being  equal,  all  the  vertical  angles  about  the  point 
o  are  equal.  But  the  angles  oef,  oaf,  oag,  obo,  made  by 
the  tangents  and  radii,  are  right  angles  ;  therefore  oef  +  oaf 
=  two  right  angles,  and  oag  +  obg  =  two  right  angles  ; 
consequently,  also,  aoe  -f  afe  =  two  right  angles,  and  aob 
-|-  AGB  =  two  right  angles  (cor.  2,  th.  18).  Hence,  then,  the 
angles  aoe  -\-  afe  being  =  aob  -f  agb,  of  which  aob  is  = 
AGE  ;  consequently  the  remaining  angles  f  and  g  are  also 
equal.  In  the  same  manner  it  is  shown,  that  all  the  angles  f, 
G,  H,  I,  K,  are  equal. 

Again,  the  tangents  from  the  same  point  fe,  fa,  are  equal, 
as  also  the  tangents  ag,  gb  (cor.  2,  th.  61)  ;  and  the  angles 
F  atid  G  of  the  isosceles  triangles  afe,  agb,  are  equal,  as  well 
as  their  opposite  sides  ae,  ab  ;  consequently  those  two  trian- 
gles are  identical  (th.  1),  and  have  their  other  sides  ef,  fa,  ag, 
GB,  all  equal,  and  fg  equal  to  the  double  of  any  one  of  them. 
In  like  manner  it  is  shown,  that  all  the  other  sides  gh,  hi,  ik, 
KF,  are  equal  to  fg,  or  double  of  the  tangents  gb,  bh,  &c. 

Hence,  then,  the  circumscribed  figure  is  both  equilateral 
and  equiangular,  which  was  to  be  shown. 

Corol, 


366  . 


GEOMETRY. 


Corol  The  inscribed  circle  touches  the  middles  of  the  sides 
of  the  polygon. 

PROBLEM  XXXII. 

To  inscribe  a  circle  in  a  Regular  Polygon. 

Bisect  any  two  sides  of  the  polygon 
by  the  perpendiculars  go,  fo,  and  their 
intersection  o  will  be  the  centre  of  the 
inscribed  circle,  and  o'g  or  of  will  be 
the  radius. 

For  the  perpendiculars  to  the  tangents 

AF,  AG,  pass  through  the  centre  (cor.  th. 

47)  ;  and  the  inscribed  circle  touches  C  D 

the  middle  points  f,  g,  by  the  last  corollary.  Also,  the  two 
sides  AG,  Ao,  of  the  right-angled  triangle  aog,  being  equal 
to  the  two  sides  af,  ao,  of  the  right-angled  triangle  aof,  the 
third  sides  of,  og,  will  also  be  equal  (cor.  th.  45).  Therefore 
the  circle  described  with  the  centre  o  and  radius  og,  will  pass 
through  F,  and  will  touch  the  sides  in  the  points  g  and  f.  And 
the  same  for  all  the  other  sides  of  the  figure. 


PROBLEM  XXXni. 


To  Describe  a  Circle  about  a  Regular  Polygon. 

Bisect  any  two  of  the  angles,  c  and  d, 
with  the  lines  co,  do  ;  then  their  inter- 
section o  will  be  the  centre  of  the.  cir- 
cumscribing circle  :  and  oc,  or  od,  will 
be  the  radius, 

For,  draw  ob,  oa,  oe,  &c.  to  the 
angular  points  of  the  given  polygon.  Then 
the  triangle  ocd  is  isosceles,  having  the  angles  at  c,  and  d 
equal,  being  the  halves  of  the  equal  angles  of  the  polygon 
BCD,  CDE  ;  therefore  their  opposite  sides  co,  do,  are  equal 
(th.  4).  But  the  two  triangles  ocd,  ocb,  having  the  two  sides 
oc,  CD,  equal  to  the  two  oc,  cb,  and  the  included  angles  ocd, 
ocb  also  equal,  will  be  identical  (th.  l),and  have  their  third 
sides  bo,  od,  equal.  In  like  manner  it  is  shown,  that  all  the 
lines  OA,  ob,  oc,  od,  oe,  are  equal.  Consequently  a  circle 
described  with  the  centre  o  and  radius  oa,  will  pass  through 
all  the  other  angular  points,  b,  c,  d,  &c.  and  will  circumscribe 
the  polygon. 


PROBLEM 


PROBLEMS. 


367 


PROBLEM  XXXIV. 

To  make  a  Square  Equal  to  the  Sum  of  two  or  more  Giveo 

Squares. 

Let  ab  and  ac  be  the  sides  of  two 
given  squares.  Draw  two  indefinite 
lines  AP,  AQ,  at  right  angles  to  each 
other ;  in  which  place  the  sides  ab, 
AC,  of  the  given  squares  ;  join  bc  ; 
then  a  square  described  on  bc  will  be 

equal  to  the  sum  of  the  two  squares  

described  on  ab  and  ac  (th.  34).  P     B  D    A 

In  the  same  manner,  a  square  may  be  made  equal  to  the 
sum  of  the  three  or  more  given  squares.  For,  if  ab,  ac,  ad, 
be  taken  as  the  sides  of  the  given  squares,  then,  making  ae  = 
Bc,  AD  =  AD,  and  drawing  de,  it  is  evident  that  the  square  on 
fiE  will  be  equal  to  the  sum  of  the  three  squares  on  ab,  ac,  ad. 
And  so  on  for  more  squares. 

PROBLEM  XXXV. 

To  make  a  Square  Equal  to  the  Diflference  of  two  Given 
Squares. 


^ 


Let  ab  and  ac,  taken  in  the  same 
straight  line,  be  equal  to  the  sides  of  the 
two  given  squares. — From  the  centre  a, 

with  the  distance  ab,  describe  a  circle  ;  

and  make  cd  perpendicular  to  ab,  meet-  A  C  B 

ing  the  circumference  in  d  :  so  shall  a  square  described  on  cd 
be  equal  to  ad^  —  ac^  ,  or  ab^  —  ac^  ,  as  required  (cor.  1 ,  th.  34). 

PROB  LEM  XXXVL 

To  make  a  Triangle  Equal  to  a  Given  Quadrangle  abcd. 

Draw  the  diagonal  ac,  and  parallel 
to  it  de,  meeting  ba  produced  at  e,  and 
join  ce  ;  then  will  the  triangle  ceb  be 
equal  to  the  given  quadrilateral  abcd. 

For,  the  two  triangles  ace,  acd,  be- 
ing on  the  same  base  ac,  and  between 
the  same  parallels  ac,  de,  are  equal  (th.  25)  ;  therefore,  if 
ABC  be  added  to  each,  it  will  make  bce  equal  to  abcd  (ax.  2). 

PROBLEM 


368 


GEOMETRY. 


PROBLEM  XXXVH. 
To  make  a  Triangle  Equal  to  a  Given  Pentagon  abcde. 

Draw  da  and  db,  and  also  ef,  cg, 
parallel  to  them,  meeting  ab  produced 
at  F  and  c  ;  then  draw  df  and  dg  ;  so 
shall  the  triangle  dfg  be  equal  to  the 
given  pentagon  abcde. 

For  the  triangle  dfa  =  dea,  and 
the  triangle  dgb  =  dcb  (th.  25)  ; 
therefore,  by  adding  dab  to  the  equals, 

the  sums  are  equal  (ax.  2),  that  is,  dab  +  daf  +  dbg  =  dab 
-f  dab  +  DBC,  or  the  triangle  dfg  =  to  the  pentagon  abcde. 


PROBLEM  XXXVUL 
To  make  a  Rectangle  Equal  to  a  Given  Triangle  abc. 


Bisect  the  base  ab  in  d  ;  then  raise  de 
and  bf  perpendicular  to  ab,  and  meeting  cf 
parallel  to  ab,  at  e  and  f  :  so  shall  df  be 
the  rectangle  equal  to  the  given  triangle 
ABC  (by  cor.  2,  th.  26). 


PROBLEM  XXXIX. 
To  make  a  Square  Equal  to  a  Giv^n  Rectangle  abcd. 


D 


JF 


Produce  one  side  ab,  till  be  be 
equal  to  the  other  side  bc.  On  ae 
as  a  diameter  describe  a  circle,  meet-  G-^. 

ing  BC  produced  at  f  :   then  will  bf 
be  the  side  of  the    square    bfgh, 

equal  to  the  given  rectangle  bd,  as  , 

required  ;  as  appears  by  cor.  th.  87,  A  Jl        BE 

and  th.  77. 


C\ 


APPLICATION 


[  369  1 


APPLICATION  OF  ALGEBRA 


GEOMETRY. 


Wk 


HEN  it  is  proposed  to  resolve  a  geometrical  problena- 
algebraically,  or  by  algebra,  it  is  proper,  in  the  first  place, 
to  draw  a  figure  that  shall  represent  the  several  parts  or  con- 
ditions of  the  problem,  and  to  suppose  that  figure  to  be  the 
true  one.  Then,  having  considered  attentively  the  nature  of 
the  problem,  the  figure  is  next  to  be  prepared  for  a  solution, 
if  necessary,  by  producing  or  drawing  such  lines  in  it  as  ap- 
pear most  conducive  to  that  end.  'J'his  done,  the  usual  sym- 
bols or  letters,  for  known  and  unknown  quantities,  are  em- 
ployed to  denote  the  several  parts  of  the  figure,  both  the 
known  and  unknown  parts,  or  as  many  of  them  as  necessary, 
as  also  such  unknown  Une  or  lines  as  may  be  easiest  found, 
whether  required  or  not.  Then  proceed  to  the  operation, 
by  observing  the  relations  that  the  several  parts  of  the  figure 
have  to  each  other  ;  from  which,  and  the  proper  theorems  ia 
the  foregoing  elements  of  geometry,  make  out  as  many  equa- 
tions independent  of  each  other,  as  there  are  unknown  quan- 
tities employed  in  them  :  the  resolution  of  which  equations,  in 
the  same  manner  as  in  arithmetical  problems,  will  determine 
the  unknown  quantities,  and  resolve  the  problem  proposed. 

As  no  general  rule  can  be  given  for  drawing  the  lines,  and 
selecting  the  fittest  quantities  to  substitute  for,  so  as  always 
to  bring  out  the  most  simple  conclusion,  because  different 
problems  require  different  modes  of  solution  ;  the  best  way  to 
gain  experience,  is  to  try  the  solution  of  the  same  problem 
in  different  ways,  and  then  apply  that  which  succeeds  best, 
to  other  cases  of  the  same  kind,  when  they  afterwards  occur. 
The  following  particular  directions,  however,  may  be  of  some 


l5^,  In  preparing  the  figure,  by  drawing  lines,  let  them  be 
either  parallel  or  perpendicular  to  other  lines  in  the  figure, 
or  so  as  to  form  sirinilar  triangles.  And  if  an  angle  be  given, 
it  will  be  proper  to  let  the  perpendicular  be  opposite  to  that 
angle,  and  to  fall  from  one  end  of  a  given  Une,  if  possible. 

Vol,  I.  •  48  H. 


370  APPLICATION  OF  ALGEBRA 

9,d,  In  selecting  the  quantities  proper  to  substitute  for, 
those  are  to  be  chosen,  whether  requiied  or  not.  tvbich  he 
nearest  the  known  or  given  parts  of  the  iigure,  and  by  nie-ans 
of  which  the  next  adjacent  parts  may  be  expressed  by  addi- 
tion and  subtraction  only^  without  using  surds. 

3c/,  When  two  lines  or  quantities  are  alike  related  to  other 
part*?  of  the  figure  or  problem,  the  best  way  is,  not  to  make 
use  of  either  of  them  separately,  but  to  substitute  for  their 
sum,  or  difference,  or  rectangle,  or  the  sum  of  their  alternate 
quotients,  or  for  some  line  or  lines,  in  the  figure,  to  which 
they  have  both  the  same  relation. 

4th,  When  the  area,  or  the  perimeter,  of  a  figure,  is  §iven» 
or  such  parts  of  it  as  have  only  a  remote  relation  to  the  parts 
required  :  it  is  sometimes  of  use  to  assume  another  figure 
similar  to  the  proposed  one,  having  one  side  equal  to  unity,  or 
some  other  known  quantity.  For  hence  the  other  parts  of  the 
figure  may  be  found,  by  the  known  proportiong  of  the  like 
sides,  or  parts,  and  so  an  equation  be  obtained.  For  exam- 
ples, take  the  following  problems. 

PROBLEM  I. 

In  a  Right-angled  Triangle^  having  given  the  Base  (3)yand  the 
Sum  of  the  Hypothenuse  and  Perpemiicular  (9)  ;  to  find 
both  these  two  Sides. 

Let  ABC  represent  the  proposed  triangle, 
right-angled  at  b.  Put  the  base  ab  =  3  =  6, 
and  the  sum  ac  -f  bc  of  the  hypothenuse 
and  perpendicular  =  9  =  5  ;  also,  let  x  de- 
note the  hypothenuse  ac,  and  y  the  perpen- . 
dicular  ec. 

Then  by  the  question      -     -     -     ^.  +  y  =  », 
and  by  theorems  34,    -     -     -     x-=y--\-b^. 
By  transpos.  y  in  the  1st  equ.  gives  x  =  s  —  y, 
This  value  of  x  substi.  in  the  2d, 

gives     ---'----       52  —  ^sy  -\-y^  =  2/^  +  b^\ 
Takingaway  2/^  on  both  sides  leaves  s2  — 2sy  =  b^. 
By  transpos.  ^sy  and  6*,  gives         s^  —  6^  =  est/, 

52—63 

Aiid  dividing  by  2s,  gives    -   -        ~  ^  =  4  =  bc. 

2s 
Hence  x  =  s  —  y       6  =  ac. 

N.  B.     In  this  solution,  and  the  following  ones,  the  nota- 
liop  is  made  by  using  as  many  unknown  letters,  x  and  y,  as 

there 


TO  GEOMETRY.  371 

there  are  unknown  sides  of  the  triangle,  a  separate  letter  for 
each  ;  in  preference  to  using  only  one  unknown  letter  for  one 
side,  and  expressing  the  other  unknown  side  in  terms  of  that 
letter  and  the  given  sum  or  difference  of  the  sides  ;  though 
this  latter  way  would  render  the  solution  shorter  ;  because  the 
former  way  gives  occasion  for  more  and  better  practice  in  re- 
ducing equations,  which  is  the  very  end  and  reason  for  which 
these  problems  are  given  at  all. 


PROBLEM  II. 


In  a  Right-angled  Triangle,  having  given  the  Hypothenuse  (5)  ; 
and  the  sum  of  the  Base  and  Perpendicular  (7)  ;  to  find  both 
these  two  Sides, 

Let  ABC   (see  last   %.)  represent   the   proposed   triangle, 
right-angled  at  b.     Put  the  given  hypothenuse  ac  =  5  =  a,  and 
the  sum  ab  -f-  bc  of  the  base  and  perpendicular  =  7  =  s;  also 
let  X  denote  the  base  ab,  and  y  the  perpendicular  bc. 
Then  by  the  question     -     -     -     x  -{-  y  =  s 

and  by  theorem  34      -     -     -     x^-\-y^=-a^ 
By  transpos.  y  in  the  1st,  gives      x  =  s  —  y 
By  substitu.  this  valu.  for  a;,  gives  s^  —  2sy  -f  ^y^  =  a^ 
By  transposing  s^ ,  gives       -     -     2^2  —  2vy  =  a^  —  52 
By  dividing  by  2,  gives        -     -     y^  —  sy  =  ia^  —  ±s3 
By  completing  the  square,  gives    y"^  — sy  •\-  \s^  =  la^  — ^52 

By  extracting  the  root,  gives        y  —  is  =  ^  la^  —  \s^ 

By  transposing  \s,  gives      -     -    y  =  \s  ±.  ^  la^  ^is^  = 

4  and  3,  the  values  oi  x  and  y, 

PROBLEM  III. 


In  a  Rectangle,  having  given  the  Diagonal  (10),  and  the  Peri- 
meter, or  Sum  of  ail  the  Four  Sides  (28)  ;  to  fmd  each  of  the 
Sides  severally. 

Let  abcd  be  the   proposed  rectangle ,;  DC 

and  put  the  diagonal   ac  =   10  =  c?.  and  '" 

half  the  perimeter  ab  -j-  bc  oir  ad  -|-  dc 
=   14  =  a  ;  also  put  one  side  ab  =  a-, 


and  the  other  side  bc  =  y.     Hence,,  by         ^  jj 

right-angled 


372 


APPLICATION  Ot^  ALGEBRA 


right-angled  triangles,     -     -     -     -     a;2  _|>  ^2  =  ^a 

And  by  the  question,      -     --     -     x  -\-y  =a 

Then  by  transposing  y  in  the  2d,  gives  x  =a  —  y 

Thisvahiesubstitutedinthe  lst,givesa?  — 2ay-\-2y^  =  d- 

Transposing  a^ ,  gives     -     -     2y^  —  2ay  =  r/2  —  a^ 

And  dividing  by  2,  gives      -     y^   —  oy  =  ^d^  —  ia^ 

By  completing  the  square,  it  is  i/^  —  ay  -{-  \a^  =  ic?2  __  1^2 


And  extracting  the  root,  gives  y  —  i^  =  ^  ^d^  —  ^a^ 


la±y/\d^^\a^  =  8 


and  transposing  ia,  gives  y 

or  6.  the  values  of  x  and  y. 

PROBLEM  IV. 

Having  given   the  Base  and   Perpendicular  of  any   Triangle  ^ 
to  find  the  Side  of  a  Square  Inscribed  in  the  Same. 
L^T  ABC  represent  the  given  triangle,  f^ 

and  EFGH  its  inscribed  square.  Put  the 
base  AB  =  b,  the  perpendicular  cd  =  a, 
and  the  side  of  the  square  gf  or  gh  =  Gr, 

Di  =1  x;  then  will  ci  =  cd  —  di   = 


Then,  because  the  like  lines  in  the 
sinoilar  triangles  abc,  gfc,  are  propor- 
tional   (by    theor.    84,    Geom.),   ab 


H     DEB 

GF  :  ci,  that 
is  b  :  a  :  :  X  :  a  —  x.  Hence,  by  multiplying  extremes  and 
means,  ab  —  fex  =  ax,  and  transposing  bx,  gives  ab  =  ax 

ab 

-}-  bx  ;  then  dividing  by    a  -\-  b,  gives  x  =  =  gf 

a  -f  6 
©r  GH  the  side  of  the  inscribed  square  :  which  therefore  is  of 
the  same  magnitude,  whatever  the  species  or  the  angles  of 
the  triangles  may  be. 

PROBLEM  V. 

In  an  Equilateral  Triangle^  havirig  given  the  lengths  of  the  three 
Perpendiculars,  drawn  from  a  certain  Point  within,  on  the  . 
ihree  Sides  :  to  determine  the  Sides. 
Let  ABC  represent  the  equilateral  tri- 
angle, and    BE,  DF,  DG,  the  given  per- 
pendiculars {ron\  the  point  d.     Draw  the 
lines  DA,  DB,   DC,  to  the   three   angular 
points  ;  and  let  fall  the  perpendicular  ch 
on  the  base  ab.  Put  the  three  given  per- 
pendiculars  DE  s=   a,    DF  =    6,    DG  =  C, 

and  put  Of  =  AH  or  bh,  half  the  side  of 


TO  GEOMETRY.  373 

the   equilateral  triangle.     Then  is  ac  or  bc  =  2r,  and  by 

right  angled  triangles  the  perpendicular  ch  =  ^  ac^  —  ah^ 

=  ^  4x2  _  x3  =  y  Sx^  =  or  ^  3. 

Now,  since  the  area  or  space  of  a  rectangle,  is  expressed 
by  the  product  of  the  base  and  height  (cor.  2,  th.  81  Geom.), 
and  that  a  triangle  is  equal  to  half  a  rectangle  of  equal  base 
and  hei^hl  (cor.  1 ,  th  26),  it  follows  that, 

the  whole  triangle  abc  is  =  ^ab  XcH=xXxy/3  =  x^^3, 
the  triangle  abd  =  Iab  X  dg  =  a;   X   c  =  cjt, 
the  triangle  bcd  =  Ibc  X  de  =  a;   X  a  =  ax, 
the  triangle  acd  =  ^ac  X  df  =  a;    X  6  =  6x. 

But  the  three  last  triangles  make  up,  or  are  equal  to,  the 
whole  former,  or  great  triangle  ; 

that  is,  x^^  3  =  ax  -{-  bx  -\- ex  ;  hence,  dividing  by  rr,  gives 
X  ^  3  =  a    4-  ^    +^>  and  dividing   by  ^3,  gives 
'\ia+b-i-cl 
X  =^ ^,  half  the  side  of  the  triangle  sought. 

Also,  since  the  whole  perpendicular  ch  is  =  ar  y^  3,  it  is 
therefore  =  a  -f  6  -j-  c.  That  is,  the  whole  perpendicular 
CH,  is  just  equal  to  the  sum  of  all  the  three  smaller  perpen- 
diculars DE  4*  DP  +  DG  taken  together,  wherever  the  point  d 
is  situated. 


PROBLEM  VI. 

In  a  Right-angled  Triangle,  having  given  the  Base  (3),  and 
the  Difference  between  the  Hypothenuse  and  Perpendicular 
(1)  ;  to  find  both  these  two  Sides. 

PROBLEM  Vn. 

In  a  Right-angled  Triangle,  having  given  the  Hypothenuse 
rs),  and  the  Difference  between  the  Base  and  Perpendicular 
(1)  ;  to  determine  both  these  two  Sides. 

PROBLEM  VUL 

•  Having  given  the  Area,  or  Measure  of  the  Space,  of  a 
Rectangle,  inscribed  in  a  given  Triapgle  j  to  determine  the 
Sides  of  the  Rectangle. 

PROBLEM 


:m  APPLICATION  OF  ALGEBRA 


PROBLEM  IX. 

In  a  Triangle,  having  given  the  Ratio  of  the  two  Sides^ 
together  with  both  the  Segments  of  the  Base,  raade  by  a  Per- 
pendicular from  the  Vertical  Angle  j  to  determine  the  Sides  of 
the  Triangle. 


PROBLEM  X. 

In  a  Triangle,  having  given  the  Base,  the  Sam  of  the  other 
two  Sides,  and  the  Length  of  a  Line  drawn  from  the  Vertical 
Angle  to  the  Middle  of  the  Base  ;  to  find  the  Sides  of  the  Tri- 
angle. 


PROBLEM  XI. 

In  a  Triangle,  having  given  the  two  Sides  about  the  Vertical 
Angle,  with  the  Line  bisecting  that  Angle,  and  terminating  in 
the  Base  :  to  find  the  Base. 


PROBLEM  XU. 

To  determine  a  Right-angled  Triangle;  having  given  the 
Lengths  of  two  Lines  drawn  from  the  acute  angles,  to  the 
Middle  of  the  opposite  Sides. 


PROBLEM  XIII. 

To  determine  a  Right- Angled  Triangle  ;  having  given  the 
Perimeter,  and  the  Radius  of  its  Inscribed  Circle, 


PROBLEM  XIV. 

To  determine  a  Triangle  ;  having  given  the  Base  the  Per- 
pendicular,  and  the  Ratio  of  the  two  Sides. 


PROBLEM  XV. 

To  determine  a  Right-angled  Triangle  ;  having  given  the 
Hypothenuse,  and  the  Side  of  the  Inscribed  Square. 

PROBLEM 


W  GEOMETRY.  37i 


iPBOBUEM  3CVI. 


To  determine  the  Radii  of  ihtee  Equal  Circles,  described 
in  a  given  Circle,  to  touch  each  other  and  also  the  Circumfer- 
ence of  the  given  Circle. 


PROBLEM  XVn. 


In  a  Right-angled  Triangle,  having  given  the  Perimeter  or 
Sum  of  all  the  Sides,  and  the  Perpendicular  let  fall  from  the 
Rii^ht  Angle  on  the  Hypdthenuse  ;  to  determine  the  1  rianglC; 
that  is,  its  Sides. 


PROBLEM  XVIIL 


To  determine  a  Right-angled  Triangle  ;  liaving  given  the 
Hypothenuse,  and  the  Difference  of  two  lines  drawn  from  the 
two  acute  angles  to  the  Centre  of  the  Inscribed  Circle. 


PROBLEM  XIX. 

To  determine  a  Triangle  ;  having  given  the  Base,  the  Per 
pendicular,  and  the  Difference  of  the  two  other  Sides. 


PROBLEM  XX, 

To  determine  a  Triangle  ;  having  given  the  Base,  the  Per- 
pendicular, and  the  Rectangle  or  Product  of  the  two  Sides. 


PROBLEM  XXI. 

To  determine  a  Triangle  ;  having  given  the  Lengths  ©f 
three  Lines  drawn  from  the  three  Angles,  to  the  Middle  of  the 
opposite  Sides. 


PROBLEM  XXII. 

In  a  Triangle,  having  given  all  the  three  Side*  ;  to  find  the 
Radius  of  the  Inscribed  Circle. 

PfiQBLfiM 


37«  APPLICATION  OF  ALGEBRA,  ko. 


PROBLEM  XXIIL 

To  determine  a  Right-angled  Triangle  ;  having  given  the 
Side  of  the  Inscribed  Square,  and  the  Radius  of  the  Inscribed 
Circle. 


PROBLEM  XXiy. 

To  determine  a  Triangle,  and  the  Radius  of  the  Inscribed 
Circle  ;  having  given  the  Lengths  of  three  Lines  drawn  from 
the  three  Angles,  to  the  Centre  of  that  Circle. 


PROBLEM  XXV. 

To  determine  a  Right-angled  Triangle  ;  having  given  the 
Hypothenuse,  and  the  Radius  of  the  Inscribed  Circle. 


PROBLEM  XXVI. 

To  determine  a  Triangle  ;  having  given  the  Base,  the  Line 
bisecting  the  Vertical  Angle,  said  the  Diameter,  of  the  Cir- 
cumscribing Circle. 


PLANE 


[     377     ] 


PLANE  TRIGONOMETRY. 


DEFINITIONS. 


1.  JT  LANE  TRIGONOMETRY  treats  of  the  rela- 
tions and  calculations  of  the  sides  and  angles  of  plane  tri- 
angles. 

2.  The  circumference  of  every  circle  (as  before  observed 
in  Geom.  Def  67)  is  supposed  to  be  divided  into  360  equal 
parts,  called  Degrees  ;  also  each  degree  into  60  Minutes, 
each  minute  into  60  Seconds,  and  so  on.  Hence  a  semicircle 
contains  180  degrees,  and  a  quadrant  90  degrees. 

3.  The  measure  of  an  angle  (Def  ^8,  Geom.)  is  an  arc  of 
any  circle  contained  between  the  two  lines  which  form  that 
angle,  the  angular  point  being  the  centre  ;  and  it  is  estimated 
by  the  number  of  degrees  contained  in  that  arc. 

Hence,  a  right  angle,  being  measured  by  a  quadrant,  or 
quarter  of  the  circle,  is  an  angle  of  90  degrees  ;  and  the 
sum  of  the  three  angles  of  every  triangle,  or  two  right  angles, 
is  equal  to  180  degrees.  Therefore,  in  a  right-angled  triangle, 
taking  one  of  the  acute  angles  from  90  degrees,  leaves  the 
other  acute  angle  ;  and  the  sum  of  two  angles,  in  any  trian- 
gle, taken  from  180  degrees,  leaves  the  third  angle  :  or  one 
angle  being  taken  from  U'O  degrees,  leaves  the  stim  of  the 
other  two  angles. 

4.  Degrees  are  marked  at  the  top  of  ihe  figure  with  a  small  **, 
minute  with  ',  seconds  with  ",  and  so  on.  Thus  57  •"  30'  12", 
denote  67  degrees  30  minutes  and  12  seconds. 


l^A^ 


5.  The  Complement  of  an  arc,  is 
what  it  wants  of  a  quadrant  or  OO**, 
Thus,  if  AD  be  a  quadrant,  then  bd  is 
the  complement  of  the  arc  ab  ;  and, 
reciprocally,  ab  is  the  complement  of 
BD.  So  that,  if  AB  be  an  arc  of  50°, 
then  its  complement  bd  will  be  40°. 

6.  The  Supplement  of  an  are,  is 
what  it  wants  of  a  semicircle,  or 
180°.  Thus,  if  ade  be  a  semicircle,  then  bde  is  the  supple- 
ment of  the  arc  ab  ;  and,  reciprocally,  ab  is  the  supplement  of 
the  arc  bde.  So  that,  if  ab  be  an  arc  of  60°,  then  its  supple- 
ment bde  will  be  130°. 


Vo*.  I. 


49 


7.  The 


378  PLANE  TRIGONOMETRY. 

7.  The  Sine,  or  Right  Sine,  of  an  arc,  is  the  line  tJrawn 
from  one  extremity  of  the  arc,  perpendicular  to  the  diameter 
which  passes  through  the  other  extremity.  Thus,  bf  is  the 
sine  of  the  arc  ab,  or  of  the  supplemental  arc  bde.  Hence 
the  sine  (bf)  is  half  the  chord  (bg)  of  the  double  arc  (bag). 

8.  The  Versed  Sine  of  an  arc,  in  the  part  of  the  diameter  in- 
tercepted between  the  arc  and  its  sine.  So,  af  is  the  versed 
sine  of  the  arc  ab,  and  ef  the  versed  sine  of  the  arc  edb. 

9.  The  Tangent  of  an  arc,  is  a  line  touching  the  circle  ia 
one  extremity  of  that  arc,  continued  from  thence  to  meet  a 
line  drawn  from  the  centre  through  the  other  extremity  ; 
which  last  line  is  called  the  Secant  of  the  same  arc.  Thus, 
AH  is  the  tangent,  and  ch  the  secant  of  the  arc  ab.  Also, 
EI  is  the  tangent,  and  ci  the  secant,  of  the  supplemental  arc 
BDE.  And  this  latter  tangent  and  secant  are  equal  to  the  for- 
mer, but  are  accounted  negative,  as  being  drawn  in  an  oppo- 
site or  contrary  direction  to  the  former. 

10.  The  Cosine,  Cotangent,  and  Cosecant,  of  an  arc, 
are  the  sine,  tangent,  and  secant  of  the  complement  of  that 
arc,  the  Co  being  only  a  contraction  of  the  word  comple- 
raent.  Thus,  the  arcs  ab,  bd,  being  the  complements  of 
each  other,  the  sine,  tangent,  or  secant  of  the  one  of  these, 
is  the  cosine,  cotangent,  or  cosecant  of  the  other.  So,  bf, 
the  sine  of  ab,  is  the  cosine  of  bd  ;  and  bk,  the  sine  of 
BD,  is  the  cosine  of  ab  :  in  like  manner  ah,  the  tangent 
of  ab,  is  the  cotangent  of  bd  ;  and  dl,  the  tangent  of  db, 
is  the  cotangent  of  ab  :  also,  ch,  the  secant  of  ab,  is  the 
cosecant  of  bd  ;  and  cl,  the   secant  of  bd,  is  the  cosecant  of 


Corol.  Hence  several  remarkable  properties  easily  follow 
from  these  definitions  ;  as, 

\st^  That  an  arc  and  its  supplement  have  the  same  sine, 
tangent,  and  secant  ;  but  the  two  latter,  the  tangent  and 
secant  are  accounted  negative  when  the  arc  is  greater  than  a 
quadrant  or  90  degrees. 

2rf,  When  the  arg  is  0,  or  nothing,  the  sine  and  tangent 
are  nothing,  but  the  secant  is  then  the  radius  ca,  the  least  it 
can  be.  As  the  arc  increases  from  0,  the  sines,  tangents, 
and  secants,  all  proceed  increasing,  till  the  arc  becomes  a 
whole  quadrant  ad,  and  then  the  sine  is  the  greatest  it  can  be, 

beinji; 


DEFINITIONS^ 


379 


being  the  radius  cd  of  the  circle  :  and  both  the  tangent  and 
secant  are  infinite. 

3d,  Of  an  arc  ab,  the  versed  sine  af,  and  cosine  bk,  or  cf, 
together  make  up  the  radius  .ca  of  the  circle.— The  radius 
CA,  the  tangent  ah,  and  the  secant  ch,  form  a  right-angled  tri- 
angle CAH.  So  also  do  the  radius,  sine,  knd  cosine,  form  ano- 
ther right  angled-triangle  cbf  or  cbk.  As  alsathe  radius,  co- 
taiiorent,  and  cosecant,  another  right-angled  triangle  cdl.  And 
all  these  right-angled  triangles  are  similar  to  each  other. 

11.  The  sine,  tangent,  or 
secant  of  an  angle,  is  the  sine, 
tangent,  or  secant  of  the  arc 
by  which  the  angle  is  measur- 
ed, or  of  the  degrees,  &c.  in 
the  same  arc  or  angle. 

12.  The  method  of  con- 
structing the  scales  of  chords, 
sines,  tangents,  and  secants, 
usu:\lly  engraven  on  instru- 
ments, for  practice,  is  exhi- 
bited io  the  annexed  figure. 

13.  A  Trigonometrical  Ca- 
non, is  a  table  showing  the 
length  of  the  sine,  tangent,  and 
secant,  to  every  degree  and  2 
minute  of  the  quadrant,  with  ^ 
respect  to  the  radius,  which 
is  expressed  by  unity  or  I, 
with  an}'  number  of  cyphers. 
The  logarithms  of  these  sines, 
tangents  and  secants,  are  also 
ranged  in  the  tables;  and  these 
are  most  commonly  used,  as  they  perform  the  calculations  by 
only  addition  and  subtraction,  instead  of  the  multiplication  and 
division  by  the  natural  sines,  &c.  according  to  the  nature  of 
logarithms.  Such  a  table  of  log.  sines  and  tangents,  as  well  as 
the  logs,  of  common  numbers,  are  placed  at  the  end  of  the  se- 
cond volume,  and  the  description  and  use  of  them  are  as  fol- 
low ;  viz.  of  the  sines  and  tangents  ;  and  the  other  table,  of 
common  logs,  has  been  alreadv  explained. 

Des- 


38i  PLANE  TRIGONOMETRY. 


Description  of  the  Table  of  Log.  Sines  and  Tangents. 

In  the  first  column  of  the  table  are  contained  all  the  arcs, 
or  angles,  for  every  minute  in  the  quadrant,  viz  from  1'  to  ' 
45**,  descending  from  top  to  bottom  by  the  left-hand  side,  and 
then  returning  back  by  the  right-hand  side,  ascending  from 
bottom  to  top,  from  45°  to  90°  ;  the  degrees  being  set  at  top 
or  bottom,  and  the  minutes  in  the  column.  Then  the  sines, 
cosines,  tangents,  cotangents,  of  the  degrees  and  minutes,  are 
placed  on  the  same  lines  with  them,  and  in  the  annexed 
columns,  according  to  their  several  respective  names  or  titles, 
which  are  at  the  top  of  the  columns  for  the  degrees  at  the  top, 
but  at  the  bottom  of  columns  for  the  degrees  at  the  bottom  of 
the  leaves.  The  secants  and  cosecants  are  omitted  in  this 
table,  because  they  are  so  easily  found  from  the  sines  and  co- 
sines ;  for,  of  every  arc  or  angle,  the  sine  and  cosecant  to- 
gether tnake  up  20  or  double  the  radius,  and  the  cosine  and 
secant  together  make  up  the  same  20  also.  Therefore,  if  a 
secant  is  wanted,  we  have  only  to  subtract  the  cosine  from  20  ; 
or,  to  find  the  cosecant,  take  the  sine  from  20.  And  the  best 
way  to  perform  these  subtractions,  because  it  may  be  done  at 
sight,  is  to  begin  at  the  left  hand,  and  take  every  figure  from 

9,  but  the  last  or  right  hand  figure  from  10,  prefixing  1,  for 

10,  before  the  first  figure  of  the  remainder. 


PROBLEM  I. 


To  compute  the  Natural  Sine  and  Cosine  of  a  Given  Arc. 

This  problem  is  resolved  after  various  ways.  One  of  these 
is  as  follows,  viz.  by  means  of  the  ratio  between  the  diameter 
and  circumference  of  a  circle,  together  with  the  known  seriei 
for  the  sine  and  cosine,  hereafter  demonstrated.  Thus,  the 
semicircumference  of  the  circle,  whose  radius  is  1,  being 
3.141592663589793  &,c.    the  proportion  will    therefore    be, 

as  the  number  of  degrees  or  minutes  in  the  semicircle, 

is  to  the  degrees  or  minutes  in  proposed  arc, 

so  is  3. 14159265  &c   to  the  length  of  the  said  arc. 

This  length  of  the  arc  being  denoted  by  the  letter  a ;  aaA 


ite 


\ 


fROBLEMS.  381 

its  s\ne  and  cosine  by  5  and  c  ;  then  will  these  two  be  express- 
ed by  the  two  follow  ing  series,  viz. 

s  =  a 1 h&c. 

2.3         2.3.4.5      2.3.4.5.6.7 
a^         a^  a 

6         120       5040 

G=;  1 { h&c. 

2       2.3.4       2.3.4.5.6 
a3       a*       a^ 

=  1 H h&c. 

2  24       720 

Exam.  1.  If  it  be  required  to  find  the  sine  and  cosine  of 
#ne  minute.  Thes  the  number  of  minutes  in  ISO**  being 
10800,  it  will  be  first,  as  10800  :  1  :  :  3  1415^266  kc.  : 
•000290888208665  =  the  length  of  an  arc  of  one  minute. 
Therefore,  in  this  case, 

a  =   0002908882 
and  ia»  =  -000000000004  &c. 
the  diflf.  is  s  =  -0002908882  the  sine  of  1  minute. 
Also,  from  1 . 

take  itt2  =  00000000423079  &c. 
leave  c  =    -9999999577  the  cosine  of  1  minute. 

Exam.  2.     For  the  sine  and  cosine  of  5  degrees. 
Here,  as  180<>  :  5°   :  :  3-14159265  &c.   :  -08726646  =  a  the 
length  of  5  degrees.     Hence  a  =  -08726646 
»«ia3  =  —  -00011076 
4._i_a5  =  '-00000004 


these  collected  give  s  =  -08715574  the  sine  of  S<*. 


And,  for  the  cosine,  1  =  1- 

__  ±a2  =^  —  -00380771 
-f._Ca4=       -00000241 

these  collected  give  c  =       -99(519470  the  cosine  of  5**. 


After  the  same  manner,  the  sine  and  cosine  of  any  other 
arc  mny  be  computed.  But  the  greater  the  arc  is  the  slower 
the  <*eries  will  converge,  iti  which  case  a  greater  number  of 
terms  must  be  taken,  to  bring  out  the  conclusion  to  the  same 
degree  of  exactness. 

Or. 


382  FLANE  TRIGONOMETRY. 

Or,  having  found  the  sine,  the  cosine  will  be  found  from 
it,  by  the  property  of  the  right-angled  triangle  cbf,  viz.  the 
cosine  of  =  y'  cb^— bf^,  or  c  =  ^  1— a-s. 

There  are  also  other  methods  of  constructing  the  canon 
of  sines  and  cosines,  which  for  brevity's  sake,  are  here 
ooutted. 


PROBLEM  II. 

To  compute  the  Tangents  md  Secants. 

The  sines  and  cosines  being  known,  or  found  by  the 
foregoing  problem  ;  the  tangents  and  secants  will  be  easily 
found,  from  the  principle  of  similar  triangles,  in  the  foUow'- 
ing  manner  : 

In  the  first  figure,  where,  of  the  arc  ab,  bf  is  the  sine, 
CF  or  BK  the  cosine,  ah  the  tangent,  gh  the  secant,  dl 
the  cotangent,  and  cl  the  cosecant,  the  radius  being  ca  or 
CB  or  CD  ;  the  three  similar  triangles  cf  b,  cah,  cdl,  give  the 
following  proportions  : 

\st,  CF  :  FB  :  :  GA  :  AH  ;  whence  the  tangent  is  known, 
being  a  fourth  proportional  to  the  cosine,  sine,  and  radius. 

^dy  cF  :  cB  :  :  ga  :  ch  ;  whence  the  secant  is  known, 
being  a  third  proportional  to  the  cosine  and  radius. 

orf,  BF  :  Fc  :  :  CD  :  DL  ;  whence  the  cotangent  is  known, 
being  a  fourth  proportional  to  the  sine,  cosine,  and  radius. 

4(h,  BF  :  BG  :  :  cd  :  cl  :  whence  the  cosecant  is  known, 
being  a  third  proportional  to  the  sine  and  radius. 

As  for  the  log.  sines,  tangents,  and  secants,  in  the  tables, 
they  are  only  the  logarithms  of  the  natural  sines,  tangents, 
and  secants,  calculated  as  above. 

HAVING  given  an  idea  of  the  calculation  and  use  of  sines, 
tangents,  and  secants,  we  may  now  proceed  to  resolve  the 
several  cases  of  Trigonometry  ;  previous  to  which,  however, 
it  may  be  proper  to  add  a  few  preparatory  notes  and  observa* 
tions,  as  below. 

-  JVote  1.  There  are  usually  three  methods  of  resolving  tri- 
angles, or  the  cases  of  trigonometry  ;  namely,  Geometrical 
Construction,  Arithmetical  Computation,  and  Instrumental 
Operation. 

In  the  first  Method^  The  triangle  is  constructed,  by  making 
the  parts  of  the  given  magnitudes,  namely,  the  sides  from  a 
scale  of  equal  parts,  and  the  angles  from  a  scale  of  chords, 


•       THEOREM  I.  333 

or  by  ?ome  other  instrument.  Then  mpasuring  the  unknown 
p?irts  by  the  same  scales  or  instruments,  the  solution  will  be 
obtained  near  the  trutlf. 

In  the  Second  Method,  Having  stated  the  term§  of  the  pro- 
portion according  to  the  proper  rule  or  theorem,  resolve  it 
like  any  other  proportion,  in  which  a  fourth  term  is  to  be 
found  from  three  given  terms,  by  multiplying  the  second 
and  third  together,  and  dividing  the  product  by  the  first,  in 
working  with  the  natural  numbers  ;  or,  in  working  with  the 
logaritbjDs.  add  the  logs,  of  the  second  and  third  terms  to- 
gether, and  from  the  sum  take  the  log.  of  the  first  terra  ; 
then  the  natural  number  answering  to  the  remainder  in  the 
fourth  t**rm  sought. 

In  the  Third  Method.  Or  Instrumentally,  as  suppose  by  the 
log.  lines  on  ^  one  side  of  the  common  two-foot  scales;  Ex- 
tend the  Compasses  from  the  first  term,  to  the  second  or 
third,  which  happens  to  be  of  the  same  kind  with  it ;  then 
that  extent  will .  reach  from  the  other  term  to  the  fourth 
term,  as  required,  taking  both  extents  towards  the"  same  end 
of  the  scale. 

JVote  2.  Every  triangle  has  six  parts,  viz.  three  sides  and 
three  angles.  And  in  every  triangle,  or  case  in  trigonometry, 
there  must  be  given  three  of  these  parts,  to  find  the  other 
three.  Also,  of  the  three  parts  that  are  given,  one  of  them 
at  least  must  be  a  side  ;  because  with  the  same  angles,^  the 
sides  may  be  greater  or  less  in  any  proportion. 

JVote  3.  All  the  cases  in  trigonometry,  may  be  comprised  in 
three  varieties  only  ;  viz. 

1st,  When  a  side  anc>  its  opposite  angTc  are  given. 

2rf,  When  two  sides  and  the  contained  angle  are  given. 

3d,  When  the  three  sides  are  given. 

For  there  cannot  possibly  be  more  than  these  three  varieties 
of  cases ;  for  each  of  which  it  will  therefore  be  proper  to 
give  a  separate  theorem,  as  follows  : 


THEOREM  I. 

When  a  Side  and  its  Opposite  Angle  are  two  of  the  Given  Parts, 

Then  the  unknown  parts  will  be  found  by  this  theorem  ; 
viz.    The  sides  of  the  triangle  have  the  same  proportion  to 
each  other,  as  the  sines  of  their  opposite  angles  have. 
That  is,  As  any  one  side. 

Is  to  the  sine  of  its  opposite  angle  ; 

So  is  any  other  side, 

To  the  sine  of  its  opposite  angle. 

Bemonstr. 


384  PLANE  TRIGONOMETRY. 

Detnonsir.  For,  let  abc  be  the  pro-  C 

po?ed  triangle,  having  ab  the  greatest 
side,  and  bc  the  least.  Take  ad  = 
Bc.  considerina:  it  as  a  radius  ;  and  let 
fall  the  perpendiculars  de,  cf,  which  j^  j^  J  J 
will  evidently  be  the  ^ines  of  the  an- 
gles A  and  B,  to  the  radius  ad  or  bc« 

Now  the  triangles  ade,  acf,  are  equiangular  ;  they  therefore 
have  their  like  sides  proportional,  namely,  ac  :  cf  :  :  a» 
or  BC  :  DE  ;  that  is,  the  side  ac  is  to  the  sine  of  its  oppos^ite 
angle  b,  as  the  side  bc  is  to  the  sine  of  its   )pposite  angle  a. 

JVote  1.  In  practice,  to  find  an  angle,  begin  the  proportion 
with  a  side  opposite  to  a  given  angle.  And  to  find  a  side, 
begin  with  an  angle  opposite  to  a  given  side. 

JVo^e  2.  An  angle  found  by  thi^  rule  is  ambiguous,  or  nn- 
eertain  whether  it  be  acute  or  obtuse,  unless  it  be  a  right 
angle,  or  unless  its  magnitude  be  such  as  to  prevent  the  ambi- 
guity ;  because  the  sine  answers  to  two  angles,  which  are 
supplements  to  each  other  ;  and  accordingly  the  geometrical 
construction  forms  two  triangles  with  the  same  parts  that  are 
given,  as  in  the  example  below  ;  and  when  there  is  no  re- 
striction or  limitation  included  in  the  question,  either  of  them 
may  be  taken.  The  number  of  degrees  in  the  table,  answer- 
ing to  the  sine,  is  the  acute  angle  ;  but  if  the  angle  be  obtuse, 
subtract  those  degrees  from  180«>,  and  the  remainder  will  be 
the  obtuse  angle.  When  a  given  angle  is  obtuse,  or  a  right  one, 
there  can  be  no  ambiguity  ;  for  then  neither  of  the  other 
angles  can  be  obtuse,  and  the  geometrical  construction  will 
form  only  one  triangle. 

EXAMPLE!. 

In  the  plane  triangle  abc, 

{  ab  346  yards 
Given  ?  bc  232  yards 

(Z  A  3'/''  20' 
Required  the  other  parts. 


I.  Geometrically. 

Draw  an  indefinite  line  ;  on  which  set  off  ab  =  34i 
from  some  convenient  scale  of  equal  parts. — Make  the  angle 
A  =  37^1. — With  a  radius  of  232,  taken  from  the  Scme 
scale  of  equal  parts,  and  centre  b,  cross  ac  in  the  two 
points  c,  G. — Lastly,  join  bc,  bc,  and  the  figure  is  con- 
structed. 


THEOREM  I.  386 

structed,  which  gives  two  triangles,  and  shows  that  the  case 
is  ambiguous. 

Then,  the  sides  ac  measured  by  the  scale  of  equal  parts, 
and  the  angles  b  and  c  measured  by  the  line  of  chords,  or 
other  instrument,  will  be  found  to  be  nearly  as  below  ;  viz. 

AC  174  Z.  B  27*'  Z.  c  115«>i. 

or  374A  or  78 a  or       64^. 


2.  Arithmetically. 

First,  to  find  the  angles  at 

c. 

As  side          bc     232 

- 

log.  2-365488 

To  sin.  op.  2L  a     37*'  20'      - 

- 

9-782796 

So  side          AB     345 

- 

2-537819 

3osin.  op.  Z.C     US*' 36'     or  640 

24'      9-955127 

add     Z.  A       37     20 

37 

20 

the  sum         152     66  or 

101 

44 

taken  from    180     00 

180 

00 

leaves  Z  b       27     04   or 

78 

16 

Then,  to  find  the  side  ac. 

As  sine        Z.  a    37*^  20' 

_ 

log.  9-782796 

To  op.  side  bc             232 

2-365488 

sosio.  .,  j-0*'    : 

9-658037 
9-990829 

To  op.  side  ac        174-07 

2-240729 

or         374-56 

2-573521 

3.  Instrumentally. 

In  the  first  proportion. — Extend  the  compasses  from  232 
to  345  on  the  Hne  of  numbers  •  then  that  extent  will  reach, 
on  the  sines,  from  37**i  to  64*>^,  the  angle  c. 

In  the  second  proportion. — Extend  the  compasses  from 
37°i  to  27<'  or  78<'i,  on  the  sines  ;  then  that  extent  will 
reach,  on  the  line  of  numbers,  from  232  to  174  or  374^, 
the  two  values  of  the  side  ac 

EXAMPLE  II. 

In  the  plane  triangle  abc, 

C      AB  366  poles  (    ^c         ^^**  ^" 

Given?    Z.A     57o  12'  Ans.  (      ac         154-33 

(    AB     24     46  (      BC         309-86 

Required  the  other  parts. 
Vol.  I.  59  E3CAMPLfi 


38e  PLANE  TRIGONOMETRY. 

EXAMPLE  III. 

In  the  plane  triangle  abg, 

5Z.B  64«  35^ 
or  116  26 
Z.C  57  57 
or  7  7 
AB  112  6  feet. 
or  16-47  feet. 

THEOREM  n. 

When  two  Sides  and  their  Contained  Jingle  are  given. 

First  add  the  two  given  sides  together,  to  get  their  sum, 
and  subtract  them,  to  get  their  difference.  Next  subtract  the 
given  angle  from  180**,  or  two  right  angles,  and  the  remainder 
will  be  the  sum  of  the  two  other  angles  ;  thon  divide  that  by 
2,  which  will  give  the  half  sum  of  the  said  unknown  angles. 
Then  say, 

As  the  sum  of  the  two  given  sides, 
Is  to  the  difference  of  the  same  ^ides  ; 
So  is.-the  tang,  of  half  the  sum  of  their  op.  angles, 
To  the  tang  of  half  the  diff,  of  the  same  angles. 
Then  add  the  half  difference  of  the  Jungles,   so  found,  to 
their  half  sum,  and   it   will  give  the  greater  angle,  and  sub- 
tracting the  same  will  leave  the  less  angle  ;  because  the  half 
sum  of  any  two  quantities,  increased  by  their  half  difference, 
gives  the  greater,  and  diminished  by  it  gives  the  l(s» 

Then  all  the  angles  being  now  known,  the  unknown  side 
will  be  found  by  the  former  theorem. 

Note.  Instead  of  the  tangent  of  the  half  sum  of  the  un- 
known angles,  in  the  third  term  of  the  proportion,  may  be 
used  the  cotangent  of  half  the  given  angle,  which  is  the  same 
thing. 

Demonst.    Let  abc  be  the  proposed 
tri^iigie,   having  the   two   given   sides 

AC,  Bc  including  the  given  angle  c. 
With  the  centre  c,  and  radius  ca, 
tiie  less  of  these  two  sides,  describe 
a  semicircle,  meeting  the  other  side 
Br  produced  in  d,  e,  and  the  un- 
known  side    AB   in    a,  g.       Join    ae, 

AD,  CG,  and   draw   df    parallel  to  ae.     ^ 

Then  be  is  the  sum  of  the  two  given  sides  ac,  cb,  or  of 
EC,  cb  ;  and  bo  is  the  difference  of  the  same  two  giv.en  sides 

/  AC, 


THEOREM  II.  387 

AC,  BC,  or  of  CD,  CB.  Also,  the  external  angle  ace,  is  equal 
to  the  given  snna  of  the  two  internal  angles  cab,  cba  ;  but  the 
ar»i»;le  ade,  at  the  circumference,  is  equal  to  half  the  angle 
ACE  dt  the  centre  :  therefore  the  same  angle  ade  is  equal  to 
h?lf  the,  given  sum  of  the  angles  cab,  cba.  Also,  the  exter- 
Dol  angle  AGC,  of  the  triangle  bcg,  is  equal  to  the  sum  of  the 
tT7'>  internal  angles  «cb,  gbc,  or  the  angle  gcb  is  equal  to  the 
'  diiV  lonce  of  the  two  angles  agc,  gbc  ;  but  the  angle  cab  is 
equal  to  the  said  angle  agc,  these  being  opposite  to  the  equal 
sides  AC,  CG  ;  and  the  angle  dab,  at  the  circumference,  is 
equal  to  half  the  angle  dcg  at  the  centre  ;  therefore  the  angle 
dab  is  equal  to  half  the  difference  of  the  two  angles  cab.  cba  ; 
of  which  it  has  been  shown  that  ade  or  cda  is  the  half  sum. 
Now  the  angle  dae,  in  a  semicircle,  is  a  right  angle,  or  ae 
is  perpendicular  to  ad  ;  and  df,  parallel  to  ae,  is  also  per- 
pendicular to  AD  :  consequently  ae  is  the  tangent  of  cda  the 
half  sum,  and  df  the  tangent  of  dab  th£  half  difference  of 
the  angles,  to  the  same  radius  ad  by  the  definition  of  a  tan- 
gent. '  but  the  tangents  ae,  df,  being  parallel,  it  will  be  as 
BE  :  bd  :  :  AE  :  DF  ;  that  is,  as  the  sum  of  the  sides  is  to  the 
difference  of  the  sides,  so  is  the  tangent  of -half  the  su-u  of 
the  opposite  angles,  to  the  tangent  of  half  their  difference. 


EXAMPLE  I. 

In  the  plane  triangle  abc, 
C  ab  346  yards 
Given  ?  ag  17  i07  yards 
(  Z.A  37»  20' 
Required  the  other  parts. 

1.  Geometrically. 

Draw  AB  =  345  from  a  scale  of  equal  parts.  Make  the 
angle  a  =  37*»  20'.  Set  off  ac  =  174  by  the  scale  of  equal 
parts.     Join  bc,  and  it  is  done. 

'  Then  the  other  parts  being  measured,  they  are  found  to  bc 
nearly  as  follows  ;  viz.  the  side  bc  232  yards,  the  angle  b  27**, 
and  the  angle  c  115*»|^. 

2.  Arithmetically. 

The  side  ab     345  From         180*>  00' 

the  side   ac     174-07  take/. a       37    20 


their  sum         519-07         sum  of  c  and  b   142    40 
their  differ.      170-93         half  sum  of  do.    71    20 


As 


388  PLANE  TRIGONOMETRY. 


As   sum  of  sides  ab,  ac,     -     -     619  07 

log.     2-716226 

To  diff.  of  sides  ab,   ac,     -     -     170-93 

2-232818 

So  tang,    half   sum   Zs  c  and  B      71'' 20' 

-      10-471298 

To  tang,  half   diff.   ^s  c  and  b      44     16 

9-988890 

these  added  give  Z.  c    116    36 

and  subtr.       give  z.  b      27      4 

Then,  by  the  former  theorem, 

As  sin.  Z.  c- 1 16°    36'  or  64«  24' 

log.  9-956126 

To  its  op.  side  ab   346 

2-637819 

Bo  sin.  of  Z.  A  370  20'          .         .         - 

9-782796 

To  its  op.  side  bc  232     '     - 

2-365489 

3.  Instrumentally. 

In  the  fifst  proportioTi. — Extend  the  compasses  from  619 
to  171,  on  the  Une  of  numbers  ;  then  that  extent  will  reach, 
on  the  tangents,  from  71** J-  (the  contrary  way,  because  the 
tangents  are  set  back  again  from  46**)  a  little  beyond  46, 
which  being  set  so  far  back  from  4:-,  falls  upon  44**^,  the 
fourth  term. 

In  the  second  proportion Extend  from  64<*i  to  37'>i  on 

the  sines  ;  then  that  extent  will  reach  on  the  numbers,  from 
346  to  232,  the  fourth  term  sought. 


EXAMPLE  It 
lu  the  plane  triangle  abc, 
Given  < 
Required  the  other  parts. 

EXAMPLE  IIL 

In  the  plane  triangle  abg, 

(    AC  120  yards 
Given  {    bc  112  yards  Ans.  {  Z.a  67*'  28 

(Zc  570  67'  (zb    64  35 

Required  the  other  parts. 

THEOREM  III. 

When  the  TJiree  Sides  of  a  Triangle  are  given. 

First,  let  fall  a  perpendicular  from  the  g^reatest  angle  on  the 
opposite  sine,  or  base,  dividing  it  into  two  segments,  and  the 
whole  triangle  into  t\»o  right-angled  triangles  :  then  the  pro- 
portion will  be, 

As 


le  piane  iriangie  abc, 

i  ab  366  poles  (    bc  309  86 

t  {  AC  154-33  Ans.  {  Zb  24*'  45' 

(Za57°12'  (Zc98       3 

(   AB  112-6 

,  ?Za  67*'  28 
(Zb    64  35 


THEOREM  Hi.  389 

As  the  base,  or  sums  of  the  segments. 

Is  to  the  sum  of  the  other  two  sides  ; 

So  is  the  diflference  of  those  sides, 

To  the  diflf.  of  the  segments  of  the  base. 
Then  take  half  this  diflference  of  the  segments,  and  add  it  to 
the  half  sum,  or  the  half  base,  for  the  greater  segment ;  and 
subtract  the  same  for  the  less  segment. 

Hence  in  each  of  the  two  right-angled  triangles,  there 
will  be  known  two  sides,  and  the  right  angle  opposite  to  one 
of  them  ;  consequently  the  other  angles  will  be  found  by  the 
iirst  theorem.  / 

Demonstr.  By  theor,  35,  Geom.  the  rectangle  of  the  sum 
and  difference  of  the  two  sides,  is  equal  to  the  rectangle  of 
the  sum  and  diflference  of  the  two  segments.  Therefore,  by 
forming  the  sides  of  these  rectangles  into  a  proportion  by  theor. 
70,  Geometry,  it  will  appear  that  the  sums  and  dififerences  are 
proportional  as  in  this  theorem. 

EXAMPLE  I. 

In  the  plane  triangle  abc, 
Given  r^!ff^y^^^« 


the  sides  >  ^^  ^^|.Q^  ^^^^^^^^ 

P     B 
To  find  the  angles. 

I.  Geometrically. 

Draw  the  base  ab  =  345  by  a  scale  of  equal  parts.  Witli 
radius  232,  and  centre  a,  describe  an  arc  ;  and  with  radius 
174,  and  centre  b,  describe  another  arc,  cutting  the  former  in 
c.     Join  AC,  Bc,  and  it  is  done. 

Then,  by  measuring  the  angles,  they  will  be  found  to  be 
nearly  as  follows,  viz. 

Za  27«>,  Zb  37oi,  and  ^c  115o|. 

2..  Arithmetically, 

Having  let  fall  the  perpendicular  cp,  it  will  be, 
As  the  base  ab  :  ac  -f-  bc  :  :  ac  —  bc  :  ap  —  bp, 
that  is,  as  345  :  406-07  :  :  67-93  :  68-18  =  ap  —  bp. 

its  half  is         -  3409 

the  half  base  is         172-50 

the  sum  of  these  is  206-59  =  ap 

and  their  diflf.  is         138-41  =  bp 

Then, 


390  PLANE  TRIGONOMETRY. 

Then,  in  the  triangle  apc,  right-angled  at  p, 


As  the  side       ac     -     • 

.     232 

log.  2-365488 

To  sin.  op.     ^p     •     • 

9()*> 

-     10-oooboo 

So  is  the  side  ap     -     - 

■     206-59    - 

2-316109 

Tosin.  op.  Zacp     -     • 

.       62^tyQ' 

-        9-949621 

Which  taken  from    - 

90    00 

leaves  the  ^^  a  27    04 
Again,  in  the  triangle  bpc,  right-angled  at  p, 


As  the  side       bc     - 

-     174-07 

-     log.  2-240724 

To  sin.  op.     ^p     - 

90° 

-     -     10-000000 

So  is  side          bp     - 

.     138-41 

-     -       2-441168 

To  sin.  op  ^^Bcp     - 

-     5S^°  40' 

-     -       9-900444 

which  tauten  from     - 

90    00 

leaves  the  Zb      37    20 

Also,  the  Zacp    62°  66' 

added  to  Zbcp    62    40 

gives  the  whole  Zacb  115    36 

So  that  all  the  three  angles  are  as  follow,  viz. 
the  Za  27°  4';  the  ^b  37*^  20' ;  the  Zc  116^36'. 

3.  Instrumentally* 

In  the  first  proportion. — Extend  the  compasses  from   345-  ,  '    J 
to  406,  on  the  line  of  numbers  ;  then  that  extent  will   reach,      •  ] 
on  the  same  line,  from  68  to  68  2  nearly,  which  is  the  diffe- 
rence of  the  segments  of  the  base. 

In  the  second  proportion.— ^Extend  from  232  to  206i.  on  the 
line  of  numbers  ;  then  that  extent  will  reach,  on  the  sines, 
from  90°  to  eS*'. 

In  the  third  proportion. — Extend  from  174  to  138^  ;  then 
that  extent  will  reach  from  90**  to  52*^1  on  the  sines. 


EXAMPLE  II. 

•     "\ 

In  the  plane 

triangle  abc. 

Given  i^-'f.^Pf' 
*u«  ^A  .  <  AC  164  33 

To  find  the  angles. 

(Za67o  j2' 
Ans.  {Zb24    46 
(Zc98    3 

EKAMPL5 

! 
i 

- 

1 

THEOREM  IV. 

EXAMPLE  m. 

In  the  plane 

triangle  abc 

f 

.ni    AB 

120 

( 

ZA 

57«» 

28' 

112-6 

Ans.  ( 

ZB 

57 

57 

f    BC 

112 

( 

Zc 

64 

35 

0  find  the  angles. 

301 


The  three  foregoing  theorems  include  all  the  cases  of  plane 
triangles,  both  right-angled  and  oblique.  But  there  are  other 
theorems  suited  to  some  particular  forms  of  triangles,  which 
are  sometimes  more  expedilious  in  their  use  than  the  general 
ones  ;  one  of  which,  as  the  case  for  which  it  serves  so  fre- 
quentljr  occurs,  may  be  here  taken,  as  follows  : 

THEOREM  IV. 

When  a  Triangle  is  Right-angled  ;  any  of  the  unknown  parts 
may  be  found  by  the  following  proportions  :  viz. 

As  radius 

Is  to  either  leg  of  the  triangle  ; 

So  is  tang,  of  its  adjacent  angle, 

To  its  opposite  leg  ; 

And  so  is  secant  of  the  same  angle, 

To  the  hypothenuse. 

Demonstr.  ab  being  the  given  leg,  in  the 
right-angled  triangle  abc  ;  with  the  centre 
A,  and  any  assumed  radius  ad,  describe  an 
arc  DE,  and  draw  df  perpendicular  to  ab,. 
or  parallel  to  bc.  Then  it  is  evident,  from 
the  definitions,  that  df  is  the  tangent,  and 
af  the  secant  of  the  arc  de,  or  of  the  an- 
gle A  which  is  measured  by  that  arc,  to  the  radius  ad.  Then, 
because  of  the  parallels  bc,  df  it  will  be,  -  -  -  -  as 
■  AD  :  AB  :  :  DF  :  bc  and  :  :  af  :  ac,  which  is  the  same  as 
the  theorem  is  in  words. 

Aote,  The  radius  is  equal,  either  to  the  sine  of  90*',  or  the 
tangent  of  45*^  ;  and  is  expressed  by  1,  in  a  table  of  natural 
sines,  or  by  10  in  the  log.  sines. 

EXAMPLE  I. 

In  the  right-angled  triangle  abc, 

Given  J  ^^  530^7'  43"  (    ^^  ^"^  ^^  ^"^  ^^' 

i.  Geometrically. 


392 


PLANE  TRIGONOMETRY. 


1.  Geometrically. 

Make  ab  =  162  equal  parts,  and  the  angle  a  =  53^  7'  48"  ; 
then  raise  the  perpendicular  bc,  meeting  ac  in  c.  So  shall 
AC  measure  270,  and  bc  216. 


2 

Arithmetically. 

As  radius 

- 

. 

log. 

10-000000 

To  leg  AB 

- 

162 

2-209515 

So  tang.  Z.  A 

- 

53°     r    48" 

- 

10- 124937 

To  leg  BC 

- 

216 

. 

2-334452 

So  secant  Z.  a 

-■ 

630    r    48" 

- 

10-221848 

To  hyp.  AC 

3. 

270 
Instrument  ally. 

2-431363 

Extend  the  compasses  from  45*»  to  53' j,  on  the  tangents. 
Then  that  extent  will  reach  from  162  to  216  on  the  line  of 
numbers. 

EXAMPLE  II. 


In  the  right-angled  triangle  abc, 
^.        ^    the  leg  AB  180  .        ^    ac  392-0146 

^*^^"  I    the  Za  62°  40'  ^^^'  I    bc  348-2464 

To  find  the  other  two  sides. 

JSTote.  There  is  sometimes  given  another  method  for  right- 
angled  triangles,  which  is  this  : 

ABC  being  such  a  triangle,  make  one 
leg  AB  radius  ;  that  is,  with  centre  a, 
and  distance  ab,  describe  an  arc  bf. 
Then  it  is  evident  that  the  other  leg  bc 
represents  the  tangent,  and  the  hypo- 
thenuse  ac  the  secant,  of  the  arc  bf,  or 
«f  the  angle  a. 

In  like  manner,  if  the  leg  bc  be  made 
radius  :  then  the  other  leg  ab  will  re- 
present  the   tangent,  and  the  hypothenuse  ac  the  secant,  of 
the  arc  bg  or  angle  c. 

But  if  the  hypothenuse  be  made  radius  ;  then  each  leg 
will  represent  the  sine  of  its  opposite  angle  ;  namely,  the  leg 
ab  the  sine  of  the  arc  ae  or  angle  c,  and  the  leg  bc  the  sine 
©f  the  arc  cd  or  angle  a. 

Then  the  general  rule  for  all  these  cases  is  this,  namely, 
that  the  sides  of  the  triangle  bear  to  each  other  the  same  pro- 
portion as  the  parts  which  they  represent. 

And  this  is  called,  Making  every  side  radius. 

JVote 


OF  HEIGHTS  AND  DISTANCES.  38^3 

Note  2.  When  there  are  given  two  sides  of  a  right-angled 
triangle,  to  find  the  third  side  ;  this  is  to  be  found  by  the 
property  of  the  squares  of  the  sides  in  theorem  34,  Geom. 
Tiz.  that  the  square  of  the  hypothenuse,  or  longest  side,  is 
equal  to  both  the  squares  of  the  two  other  sides  together. 
Therefore,  to  find  the  longest  side,  add  the  squares  of  the 
two  shorter  sides  together,  and  extract  the  square  root  of 
that  sum  ;  but  to  find  one  of  the  shorter  sides,  subtract  the 
one  square  from  the  other,  and  extract  the  root  of  the  re- 
mainder. 


OF  HEIGHTS  AND  DISTANCES,  &c. 

BY  the  mensuration  aad  protraction  of  lines  and  angles, 
are  determined  the  lengths,  heights,  depths,  and  distances  of, 
bodies  or  objects. 

Accessible  lines  are  measured  by  applying  to  them  some 
certain  measure  a  number  of  times,  as  an  inch,  or  foot,  or 
yard.  But  inaccessible  lines  must  be  measured  by  taking 
angles,  or  by  such-like  method,  drawn  from  the  principles  of 
geometry. 

When  instruments  are  used  for  taking  the  magnitude  of 
Ihe  angles  in  degrees,  the  lines  are  then  calculated  by  trigo- 
ndmetry  :  in  the  other  methods,  the  lines  are  calculated  from 
the  principle  of  similar  triangles,  or  some  other  geometrical 
property,  without  regard  to  the  measure  of  the  angles. 

Angles  of  elevation,  or  of  depression,  are  usually  taken 
cither  with  a  theodolite,  or  with  a  quadrant,  divided  into 
degrees  and  minutes,  and  furnished  with  a  plummet  suspend- 
ed from  the  centre,  and  two  open  sights  fixed  on  one  of  the 
radii,  or  else  with  telescopic  sights. 

To  take  an  Angle  of  Altitude  and  Depression  with  the  (Quadrant, 

Let  A  be  any  object,  as  the  sun, 
moon,  or  a  star,  or  the  top  of  a 
tower,  or  hill,  or  pther  eminence  : 
and  let  it  be  required  to  find  the 
measure  of  the  angle,  abc,  which  a 
line  drawn  from  the  object  makes 
above  the  horizontal  line  bc. 

Place  the  centre  of  the  quadrant 
in   the  angular  point,  and  move  it 

V«L.  I.  51  r»uni 


394  ;    OF  HEIGHTS 

found  there  as  a  centre,  till  with  one  eye  at  b,  the  other 
being  shut,  you  perceive  the  object  a  throngb  the  sights  ; 
then  will  tlie  arc  gh  of  the  quadrant,  cut  off  by  the  plumb- 
line  BH,  be  the  measure  of  the  angle  abc  as  required. 

The  angle  abc  of  depression  of  ^j  .   ^  ^  . 

any  object  a,  below  the  horizontal 
line  Bc,  is  taken  in  the  same  man- 


^ 


ner ;  except  that  here  the  eye  is  ap- 
plied to  the  centre,  and  the  measure  ^ 
of  the  angle  is  the  arc  gh,  on  the 
other  side  of  the  plumb-line. 


\^ 


The  following  examples  are  to  be  constructed  and  calcu- 
lated by  the  foregoing  methods,  treated  of  in  Trigonometry. 

EXAMPLE  I. 

Having  measured  a  distance  of  200  feet,  in  a  direct  ho- 
rizontal line,  from  the  bottom  of  a  steeple,  the  angle  of 
elevation  of  its  top,  taken  at  that  distance,  was  found  to  be 
47°  30'  ;  from  hence  it  is  required  to  find  the  height  of  the 
steeple.^ 

Construction, 

,  Draw  an  indefinite  line  ;  on  which  set  off  ac  =  200  equal 
parts  for  the  measured  distance.  Erect  the  indefinite  per- 
pendicular AB  ;  and  draw  cb  so  as  to  make  the  angle  c  = 
4T*  30*,  the  angle  of  elevation  ;  and  it  is  done.  Then  ab, 
measured  on  the  scale  of  equal  parts,  is  nearly  218^. 

Calculation. 

As  radius         -         -  30*000000 

To  AC  200       -         -  2-30H)30  / 

So  tang.  Z  c  47°  30'  10  037948  / 

To  AB  218-26  required  2-338978 


EXAMPLE  IL 

What  was  the  perpendicular  height  of  a  cloud,  or  of  a 
balloon,  when  its  angles  of  elevation  wore  35*^  and  64°,  as 
taken  by  two  observers,  at  the  same  tim§,  both  on  the  same 
si/le  of  it,  and  in  the  same  vertical  plane  ;  the  distance  be- 
tween them  being  half  a  mile  or  880  yards.  And  what  wa» 
its  distance  from  the  said  two  observers  ? 

Construction  i 


AND  DISTANCES. 


396 


Construction. 

Draw  an  indefinite  ground  line,  on  which  set  off  the  given 
distance  ab  =  880  :  then  a  and  b  are  the  places  of  the  ob- 
servers. Make  the  angle  a  =  35*>,  and  the  angle  b  =  h4^  ; 
then  the  intersection  of  the  lines  at  c  will  be  the  place  of  the 
balloon  :  whence  the  perpendicular  cd,  being  let  fall,  will  be 
its  perpendicular  height.  Then  by  measurement  are  found 
the  distances  and  height  nearly  as  follow,  viz.  ac  1631,  eg 
1041,  Dfc  936. 


•        Calculation. 

First,  froro^  b        64» 
take     Z  A        35 
lea>ves  ^  acb    29 


^  / 
y^    /       ' 


Then  in  the  triangle  abc, 
As  sin.  /,  acb  29** 

To  op.  side  ab         880 
So  sin.  Z  A  33^         - 

To  op.  side  bc    1041-125     - 


B 


I) 


As  sin.  ^  acb            29<*  - 

To  op.  side  ab            880  - 
So  sin.  Z  b  116®  or' 64*^ 

To  op.  side  AC     1631-442  - 

And  in  the  triangle  bcd. 

As  sin.  Z  D               90°  - 

To  op.  side  bc     1041-125  - 
So  sin.  Z  b                64° 

To  op.  side  cd      935-757  - 


^'685571 
^•944483 
9-758591 
3017503 

9-685571 
2-944^83 
9-953660 
3-212572 


10-000000 
3  017503 
9-953660 
2-971163 


EXAMPLE  III. 


Having  to  find  the  height  of  an  obelisk  standing  oij,  the  top 
of  a  declivity,  I  first  measured  from  its  bottom^  distance 
of  40  feet,  and  there  found  the  angle,  formed  by  the  oblique 
plane  and  a  Hne  imagined  to  go  to  the  top  of  the  obelisk,  41°; 
but  after  measuring  on  in  the  same  direction  60  feet  farther', 
the  like  angle  was  only  23»  45'.  What  then  was  the  height 
«f  the  obelisk  ? 

*  Construction , 


39fi 


«F  HEIGHTS 


Construction. 


Draw  an  inde6nite  line  for  the  sloping  plane  or  declitity, 
in  which  Assume  any  point  a  for  the  bottom  of  the  obelisk, 
from  which  set  off  the  distance  ac  =  40,  and  again  cd  =  60 
equal  parts.  Then  make  the  angle  c  =  41°,  and  the  angle  p 
=  23°  45' ;  and  the  point  b  where  the  two  lines  meet  will 
be  the  top  of  the  obelisk.  Therefore  ab,  joined,  will  be  its 
height. 

Calculation '  *  r^l^ 


From  the  Zc         41°  00' 
take  the    Z  »         23    45 
leaves  theZ  DBG     17    15 

DBC, 
IC, 

121-4 
41-'^ 

69^ 

42 

T 

Then  in  the  triangle  ] 

As  sin  Z  HBc  17^   15' 
To  op.  side  DC  60     - 
So  sin.  Z  D       23    45 
To  op.  side  CB  81-488 

D 

88 
188 
30' 
24i       -  . 

1 

9-472086 
i-778151 
9-605032 
l-9110qx 

And  in  the  triangle  ab 
As  sum  of  sides  cb,  ca 
To  diflf.  of  sides  cb,  ca 
So  tang,  half  sum  Zs  a,  b 
To  tang,  halfdifif  Zs  a,  b 

2-084533 

1-617923 

10-427262 

9-960652 

the  diff.  of  these  is  Z.  cba  27 

'^\ 

Lastly,  as  sin.ZcB  a  27°  b% 
To  op  side  ca             40 
So  sin.  Z  c        -         41°   0' 
To  op.  side  AB             57-623 

-    "     - 

9-658284 
1-602060 
9-816943 
1-760719 

EXAMPLE  IV. 


Wanting  to  know  the  distance  between  two  inaccessible 
trees,  or  other  objects,  from  the  top  of  a  tower  120  feet 
high,  which  lay  in  the  same  right  line  with  the  two  objects, 
I  took  the  angles  formed  by  the  perpendicular  wall  and  hnes 
conceived  to  be  drawn  from  the  top  of  the  tower  to  the  bot- 
tom of  each  tree,  and  found  them  to  be  33°  and  64°.  What 
then  may  be  the  distance  between  the  two  objects^ 

Conztfuction. 


AND  DISTANCES. 


397 


Construction. 


Draw  the  indefinite  ground  line 
BD,  and  perpendicular  to  it  ba  = 
120  equal  parts.  Then  draw  the 
two  lines  ac,  ad,  making  the  two 
angles  bag,  bad,  equal  to  the  given 
angles  SS^  and  64^^.  So  shall  c 
and  D  be  the  places  of  the  two  ob- 
jects. 

Calculation. 


'm:^- 


\, 


First,  in  the  right- angled  triangle  abc. 

As  radius 10-OOOOOi 

ToAB  -       120      -         -         -  2079181 

So  tang.  Z   BAG       33«>     -  ,      -         -  9-812517 


TOBC 


77-929 


1-891698 


Then  in  the  right-angled  triangle  abd, 

As  radius         -----  10  000000 

ToAB     -         -         -      120         -         -  2-079181 

So  tang.  Z  BAD              64^       -         -  10-321504 

Toed     -         -       251-585         -         -  2-400685 
From  which  take  bo  77-929 
leaves  the  dist.  cd  1  $'3-656  as  required. 

EXAMPLE  V. 

Being  on  the  side  of  a  river,  and  wanting  to  know  the  dis- 
tance to  a  house  which  was  seen  on  the  other  side,  1  measur- 
ed 200  yards  in  a  strait  line  by  the  side  of  the  river  ;  and 
then,  at  each  end  of  this  line  of  distance,  took  the  horizontal 
angle  formed  between  the  house  and  the  other  end  of  the  line  ; 
which  angles  were,  the  one  of  them  68^  2',  and  the  other 
73<>  15'.  What  then  were  the  distances  from  each  end  to  the 
house  ? 

Construction. 


Draw  the  line  ab  =  200  equal  parts.  Then  draw  ac  so  as 
to  make  the  angle  a  =  68°  2',  and  bc  to  make  the  angle  b  = 
730  15'.  So  shall  the  point  c  be  the  place  of  the  house  re- 
quired. 

Calculation, 


398  OF  HEIGHTS 

Calculation. 

To  the  given  ^  a  68**  2' 
add  the  given  ^  b  73  15 
then  their  sum  141     17 

being  taken  from         1 80       0 

leaves  the  third  ^q      38     43  

A  B     , 

Hence,  As  sin.  ^  c  38«»         43'         -         9-796206  - 

To  op.  side  AB  200  -         2-301030 

So  sin.   Z  A                  68<»  2'         -         9-967268 

To  op.  side  Bc  296  64,  -         2-472092 

And,  As  sin.  Z  c  3?**  43'  -  -         9-796206 

To  op.  side  AB  200  -         2-301030 

So  sin  Z  B  73P  15'  -         9-981171 

To  op.  side  AC  306-19  -         2-485996 

Exam.,  vi.  From  the  edge  of  a  ditch,  of  36  feet  wide,  sur- 
rounding a  fort,  having  taken  the  angle  of  elevation  of  the  top 
of  the  wall,  it  was  found  to  be  62<*  40'  :  required  the  height 
of  the  wall,  and  the  length  of  a  ladder  to  reach  from  my  sta- 
tion to  the  top  of  it  ?  .  (  height  of  wall  69-64, 
^°^-  gladder,  78-4  feet. 

Exam.  vif.  Required  the  length  of  a  shoar,  which  being  to 
strut  11  feet  from  the  upright  of  a  building,  will  support  a 
jamb  23  feet  10  inches  from  the  ground  ? 

Ans.  26  feet  3  inches. 

Exam.  viii.  A  ladder,  40  feet  long,  can  be  so  planted,  that 
it  shall  reach  a  window  33  feet  from  the  ground,  on  one  side 
of  the  street ;  and  by  turning  it  over,  without  moving  the  foot 
out  of  its  place,  it  will  do  the  same  by  a  window  21  feet  high, 
on  the  other  side  :  required  the  breadth  of  the  street  ? 

Ans.  5b-649  feet. 

Exam.  ix.  A  maypole,  whose  top  was  broken  off  by  a  blast 
of  wind,  struck  the  ground  at  15  feet  distance  from  the  foot- of 
the  pole  :  what  was  the  height  of  the  whole  maypole,  suppos- 
ing the  broken  piece  to  measure  39  feet  in  length  ? 

Ans.  75  feet. 

Exam.  x.  At  170  feet  distance  from  the  bottom  of  a  tower, 
the  angle  of  its  elevation  was  found  to  be  62°  3'  :  required 
the  altitude  of  the  tower  ?  Ans.  221-55  feet. 

Exam.  xi.  From  the  top  of  a  tower,  by  the  sea-side,  of  143 
feet  high,  it  was  observed  that  the  angle  of  depression  of  a 
ship's  bottom,  then  at  anchor,  measured  35*^  ;  what  then  was 
the  ship's  distance  from  the  bottom  of  the  wall  '^ 

Ans.  204-22  feet. 
Exam. 


AND  DISTANCES.  3Sa 

Exam.  xii.  What  is  the  perpendicular  height  of  a  hill  ? 
its  angle  of  elevation,  taken  at  the  bottom  of  it,  being  46**, 
and  200  yards  farther  ofif,  on  a  level  with  the  bottom,  the  angle 
was  SI**  ?    .  Ans.  286-28  yards. 

Exam.  xiii.  Wanting  to  know  the  height  of  an  inacces- 
sible tower  ;  at  the  least  distance  from  it,  on  the  same  hori- 
zontal plane,  I  took  its  angle  of  elevation  equal  to  58"  ; 
then  going  300  feet  directly  from  it,  found  the  angle  there  to 
be  only  32**  ;  required  its  height^,  and  my  distance  from  it  at 
the  iirst  station  ?  .       5  height       307-53 

^"^-  ^  distance     192-15 

Exam.  xiv.  Being  on  a  horizontal  plane,  and  wanting  to 
know  the  height  of  a  tower  placed  on  the  top  of  an  inac- 
cessible hill  ;  I  took  the  angle  of  elevation  of  the  top  of  the 
hill  40**,  and  of  the  top  of  the  tower  51°  ;  then  measuring  in 
a  line  directly  from  it  to  the  distance  of  200  feet  farther,  I 
found  the  angle  to  the  top  of  the  tower  to  be  23**  45'.  W^hat 
then  is  the  height  of  the  tower  ?  Ans.  9333148  feet 

Exam.  xv.  From  a  window  near  the  bottom  of  a  house, 
which  seemed  te  be  on  a  level  with  the  bottom  of  a  steeple, 
I  took  the  angle  of  elevation  of  the  top  of  the  steeple  equal 
40**  :  then  from  another  wmdow,  18  feet  directly  above  the 
former,  the  like  angle  was  37«  30'  :  what  then  is  the  height 
and  distance  of  the  steeple  ?  *        ^  height       21044 

^^^'   ^distance    250-79 

Exam.  xvi.  Wanting  to  know  the  height  of,  and  my  dis- 
tance from,  an  object  on  the  other  side  of  a  river,  which 
seemed  to  be  on  a  level  with  the  place  where  I  stood,  close 
by  the  side  of  the  river  ;  and  not  having  room  to  measure 
backward,  on  the  satne  plane,  because  of  the  immediate  rise 
of  the  bank,  I  placed  a  mark  where  I  stood,  and  measured 
in  a  direction  from  the  object,  up  the  ascending  ground  to 
the  distance  of  264  feet,  where  it  was  evident  that  I  was 
above  the  level  of  the  top  of  the  object  ;  there  the  angles 
of  depression  were  found  to  be,  viz.  of  the  mark  left  at  the 
river's  side  42*>,  of  the  bottom  of  the  object,  27*>,  and  of  its 
top  19*.  Required  then  the  height  of  the  object,  and  the 
distance  of  the  mark  from  its  bottom  ? 

^        ^height        67-26 
^"^'   ^distance   150-50 

Exam.  xvir.  If  the  height  of  the  mountain  called  the 
Peak  of  Teneriffe  be  2^  miles,  as  it  is  nearly,  and  the  angle 

take* 


400  OF  HEIGHTS 

taken  at  the  top  of  it,  as  formed  between  a  plumb-line  and  a 
line  conceived  to  to  touch  the  earth  in  the  horizon,  or  farthest 
visible  point,  be  87®  58'  ;  it  is  required  from  these  to  deter- 
mine the  magnitude  of  the  whole  earth,  and  the  utmost  dis- 
tance that  can  be  seen  on  its  surface  from  the  top  of  the  moun- 
tain, supposing  the  form  of  the  earth  to  be  perfectly  round  ? 


.         i  dist.      140-876  >     ., 
A°^-   Idiam.        IQ-Sel"^'^^'' 


Exam,  xviii.  Two  ships  of  war,  intending  to  cannonade 
a  fort,  are,  by  the  shallowness  of  the  water,  kept  so  far 
from  it,  that  they  suspect  their  guns  cannot  reach  it  with 
effect.  In  order  therefore  to  measure  the  distance,  they  se- 
parate from  6ach  other  a  quarter  of  a  mile,'  or  440  yards  ; 
then  each  ship  observes  and  measures  the  angle  which  the 
other  ship  and  the  fort  subtends,  which  angles  are  83*"  45 
and  85°  15'.  What  then  is  the  distance  between  each  ship 
and  the  fort?  .        J  2292-26  yards. 

^^^'   I  2298-05. 

Exam.  xix.  Being  on  the  side  of  a  river,  and  wanting  to 
know  the  distance  to  a  house  which  was  seen  at  a  distance  on 
the  other  side  ;  I  measured  out  for  a  base  400  yards  in  a 
right  line  by  the  side  of  the  river,  and  found  that  the"  two 
angles,  one  at  each  end  of  this  line,  subtended  by  the  other 
end  and  the  house,  were  68«*  2'  and  7?><*  15.  What  then  was 
the  distance  between  each  station  and  the  house  ? 

.        i  593-08   yards. 
^"^'   I  612-38 

Exam.  xx.  Wanting  to  know  the  breadth  of  a  river,  I 
measured  a  base  of  500  yards  in  a  straight  line  close  by  one 
side  of  it  ;  and  at  each  end  of  this  hne  I  found  the  angles 
subtended  by  the  other  end  and  a  tree  close  to  the  bank  on 
the  other  side  of  the  river,  to  be  53<'  and  79^  12'.  What  then 
was  the  perpendicular  breadth  of  the  river  ? 

Ans.  529-48  yards. 

Exam.  xxi.  Wanting  to  know  the  extent  of  a  piece  of 
water,  or  distance  between  two  headlands;  I  measured  from 
each  of  them  to  a  certain  point  inland,  and  found  the  two  dis- 
tances to  be  735  yards  and  840  yards  ;  also  the  horizontal 
angle  subtended  between  these  two  lines  was  S5<^  40'.  What 
then  was  the  distance  required  ?  Ans.    741  2  yards. 

Exam.  xxii.  A  point  of  land  was  observed,  by  a  ship  at 
sea,  to  bear  east-by-south  ;  and  after  sailing  north-east  12 
miles,  it  was  found  to  bear  south-east-by-east.     It  is  required 

t9 


AND  DISTANCES.  401 

t«  determine  the  place  of  that  headland,  and  the  ship's  dis- 
tance from  it  at  the  last  observation  ?         Ans.  260728  miles. 

Exam,  xxiii.  Wanting  to  know  the  distance  between  a 
house  and  a  mill,  which  were  seen  at  a  distance  on  the  other 
side  of  a  river,  I  measured  a  base  line  along  the  side  where 
I  was,  of  600  yards,  and  at  each  end  of  it  took  the  angles 
subtended  by  the  othe  end  and  the  house  and  mill  which 
were  as  follow,  viz  at  one  end  the  angles  were  58*^  20'  and 
95°  20',  and  at  the  other  end  the  like  angles  were  53^  30'  and 
98<*  45'.  What  then  was  the  distance  between  the  house  and 
Boill  ?  Ans.  959-5866  yard«. 

Exam.  xxiv.  Wanting  to  know  my  distance  from  an  in- 
accessible object  0,  on  the  other  side  of  a  river  ;  and  having 
no  instrument  for  taking  angles,  but  only  a  chain  or  cord  for 
measuring  distances  ;  from  each  of  two  stations,  a  and  b,  which 
were  taken  at  500  yards  asunder,  I  measured  in  a  direct  line 
from^;he  objectO  100  yards,  viz.  ac  and  bd  each  equal  to  100 
yards  ;  also  the  diagonal  ad  measured  550  yards,  and  the  dia- 
gonal Bc  560.  What  then  was  the  distance  of  the  object  0 
from  each  station  a  and  b  ?  .        i    ao  536-25 

^^^'  }    BO  500-09 

Exam.  xxv.  In  a  garrison  besieged  are  three  remarkable 
objects,  A,  B,  c,  the  distances  of  which  from  each  other  are 
discovered  by  means  of  a  map  of  the  place,  and  are  as  fol- 
low, viz.  AB  2661,  AC  530,  bc  327i  yards.  Now,  having  to 
erect  a  battery  against  it,  at  a  certain  spot  without  the  place, 
and  being  desirous  to  know  whether  the  distances  from  the 
three  objects  be  such,  as  that  they  may  from  thence  be  bat- 
tered with  effect,  I  took,  with  an  instrument,  the  horizontal 
angles  subtended  by  these  objects  from  my  station  s,  and 
found  them  to  be  as  follow,  viz.  the  angle  asb  13**  30',  and 
the  angle  bsc  29°  50'  ;  required  the  three  distances,  sa,  sb,  sc  ; 
the  object  b  being  situated  nearest  to  me,  and  between  the 
two  others  A  and  c  ?  C  sa  757-14 

Ans.  ?  SB  537-10 
I  sc  655-30 
Exam.  xxvr.  Required  the  same  as  in  the  last  example, 
when  the  object  b  is  the  farthest  from  my  station,  but  still 
seen  between  the  two  others  as  to  angular  position,  and  those 
angles  being  thus,  the  angle  asb  33*^  45',  and  bsc  22°  30', 
also  the  tliree  distances,  as  600,  ac  80U,  bc  400  yards  ? 

C    sa     709J- 
Ans.  ?    SB   1042f 
f    sc     934 
Vol.  F.  B2  MENSI7R4TION 


[  402  J 


MENSURATION  OF  PLANES. 


THE  Area  of  any  plane  figure,  is  the  measure  of  the  gpa<ie 
contained  within  its  extremes  or  bounds  ;  without  any  regard 
to  thickness. 

This  area,  or  the  content  of  the  plane  figure,  is  estimated 
by  the  number  of  little  squares  that  may  be  contained  in  it  ; 
the  side  of  those  little  measuring  squares  being  an  inrh,  a 
foot,  a  yard,  or  any  other  fixed  quantity.  And  hence  tbe  area 
or  content  is  said  to  be  so  many  square  inches,  or  square  feet, 
or  square  yards,  fitc. 


Thus,  if  the  figure  to  be  measured  be 
the  rectangle  abcd,  and  the  little  square 
E,  whose  side  is  one  inch,  be  the  mea- 
suring unit  proposed  :  then  as  often  as 
the  said  little  square  is  contained  in  the 
rectangle,  so  many  square  inches  the 
rectangle  is  said  to  contain  which  in 
the  present  c&se  is  12. 


D 

4 

C 

3 

B 


PROBLEM  I. 


To  find  the  Area  of  any  Parallelogram  ;  •whether  it  heeLSquart^ 
a  Rectangle,  a  Rhombus,  or  a  Rhomboid. 

^  ■  d ;, 

Multiply  the  length    by    the  perpendicular  breadth,   or 
height^  and  the  product  will  be  the  area*. 

EXAMPLES. 


»  Tbe  truth  of  this  rule  is  proved  in  the  Geom.  theor.  81,  cor.  2. 

The  same  is  othcwise  proved  thu-  •.  L^-.i  tf.e  foregoing  rectangle 
be  the  figure  proposed  :  «nd  let  the  1  .ngth  and  bieadth  be  divided  h^to 
several  parts  each  equal  to  the  linear  measuring  unit,  being;  here  4  for 
the  leng'h,  and  o  for  the  breadth  ;  and  kt  the  opposite  points  of  divi- 
sion be  connected  by  right  lines — Then  it  is  evident  that  these  lines 
divide  the  rectangle  into  y  niimber  of  little  squares,  ench  equal  to  the 
square  measuring  umt  e  j  and  further,  that  the  number  of  tbes^e  little 
squares,  or  ihe  aiea  of  tbe  figure,  is  equal  to  the  number  of  linear 
measuring  units  in  the  lengvh,  repealed  as  often  as  there  are  hnear 

measuring 


MENSURATION  OF  PLANES.  403 


EXAMPLES. 

Ex.  1      To  and  the  area  of  a  parallelogram,  the  length  be- 
ing 12-26  and  height  8-5. 

12-25  length 
8-5  breadth 


6126 
9800 


104-125  area, 

Ex.  2.  To  find  the  area  of  a  square,  whose  side  is  35-25 
chains.  Ans.  124  acres,  1  rood,  1  perch. 

Ex.  3.  To  find  the  area  of  a  rectangular  board,  whose 
length  is  12^  feet,  and  breadth  9  in€he8.  Ans.  9|  feet. 

Ex.  4.  To  find  the  content  of  a  piece  of  land,  in  form  of 
a  rhombus,  its  length  being  6-20  chains,  and  perpendicular 
height  5-45  Ans.  3  acres,  1  rood,  20  pp.rches. 

Ex.  6.  To  find  the  number  of  square  yards  of  painting  in 
a  rhomboid,  whose  length  is  37  feet,  and  breadth  5  fettt  3 
inches.  Ans.  21 /^square yards 


PROBLEM  II. 


Tojind  the  Area  of  a  Triangle. 

Rule  I.  Multiply  the  base  by  the  perpendicular  height,  and 
take  half  the  product  for  the  area*.  Or,  multiply  the  one  of 
these  dimensions  by  half  the  other. 


measuring  units  in  the  breadth,  or  height ;  that  is,  equal  to  the. 
length  drawn  into  the  height  ;  which  here  Is  4  X   3  or  12- 

And  it  is  proved,  (Geom.  theop.  25,  cor.  2)  that  any  oblique 
parallelogram  is  equal  to  a  rectangle,  of  equal  length  and  perpendi> 
cula>  breadth.  I'herefbre  the  rule  is  general  for  all  parallelograms 
whatever. 

♦  The  truth  of  this  rule  is  evident,  he<  use  any  triangle  is  the 
half  of  a  parallelogram  of  equal  base  and  altitude,  by  Geom.  theor. 
26. 

examples; 


404  MENSURATION 

EXAMPLES. 

Ex.  1.  To  find  the  area  of  a  triangle,  whose  base  is  626 ji 
and  perpendicular  height  520  hnks  ?  ^ 

Here  625  X  260  =  162500  square  links, 

or  equal  1  acre,  2  roods,  20  perches,  the  answer. 
Ex.  2.  How  many  square  yards  contains  the  triangle,  whose 
base  is  40,  and  perpendicular  30  feet  ? 

Ans.  66|  square  yards. 
Ex.  3.  To  find  the  number  of  square  yards  in  a  triangle, 
whose  base  is  49  feet,  and  height  25^  feet  ? 

An^.  684f,  or  68-7361. 
Ex.  4.  To  find  the  area  of  a  triangle,  whose  base  is  18  feet 
4  inches,  and  height  11  feet  10  inches  ? 

Ans.  108  feet,  5|  inches^. 

Rule  11    When  two   sides  and  their  contained  angle  are 

given  :  Multiply  the  two  given  sides  together,  and  take  half 

their  product.     Then  say,  as  radius  is  to  the  sine  of  the  given 

angle,  so  is  that  half  product,  to  the  area  of  the  triangle 

Or,  multiply  that  half  product  by  the  natural  sine  of  the  said 
angle,  for  the  area*. 

Ex.  I.  What  is  the  area  of  a  triangle,  whose  two  sides  are 
39  and  40,  and  their  contained  angle  28*^  57'  ? 

By  Natural  Numbers .  By  Logarithms. 
First  4  X  40  X  30  =  600,          \ 

then  1  :  600  :  :  -484046  sin.  28^  57'  log   9-684887 

600  2  778151 


Answer     290-4276  the  area  answering  2-463038 

Ex.  2.  How  many  square  yards  contains  the  triangle,  of 
which  one  angle  is  45°,  and  its  containing  sides  25  and  ^MA, 
fe^t  1  Ans.  20-86947. 


•  For,  let  AB,  AC,  be  the  two  given  sides, 
including  the  given  angle  A.  N'^w  ^  ab  X 
CP  is  the  area,  by  the  first  rule,  CP  being  the 
perpendicular.  But,  by  trigonometry,  as  sin. 
^  p.  or  radius  :  AC  : :  sin.  /  A  :  CP,  which  is 
therefore  «=  AC  X  sin.  Z  A,  tak.ng  radius=  1. 
Therefore  the  area  ij  aB^cp  is  =  ^  AB  X  ac 
X  sin.  Z  A,  to  radius  1  ;  or  as  radius:  sin.  jt, 
^4.  : :  ^  AB  ;<AC  :  the  area. 


P   B 


RULE  III. 


6F  PLANES. 


4t5 


Rule  III.  When  the  three  sides  are  given  :  Add  all  the 
three  sides  together,  and  take  half  that  sum.  Next,  subtract 
each  side  severally  from  the  said  half  sum,  obtaining  three 
remainders.  Then  multipl}'  the  said  half  sum  and  those  three 
remainders  all  together,  and  extract  the  square  root  of  the 
last  product,  for  the  area  of  the  triangle*. 


*  For  let  ABC  be  the  given 
trinng-le.  D  aw  the  parallels 
AE.  BD,  meeting  the  two  sides 
AC,  CB,  produced,  in  D  and  e, 
an',  making  CD  =  cb,  and 
CE  =««  CA.  Also  draw  CFo  bi- 
sect n^  DB  and  AE  perpendicu. 
laily  in  F  and  g  ;  and  fhi  pa- 
ranel  to  the  side  ab,  meeting 
AC  in  H,  and  ae  produced  in  i. 
Lastly,  with  centre  h,  and  radi- 
us hf,  describe  a  circle  meeting 

AC  produced  in  k  ;  which  will  pass  through  o,  because  c  is  a  right 
angle,  and  tfirough  i,  becausci  by  means  of  the  parallels,  Ai  =  fb  ==• 
DF,  therefire  hd  =  ha,  and  hf  =  hi"*  |ab. 

Hence  ha  or  hd  is  half  the  difference  of  the  sides  AC,  CB,  and 
HC  *=  half  their  sum  or  =  ^ac  +  ^cb  ;  also  hk  «^  hi  =  ^if  op 
^AB  ;  conseq.  ck  «=*  ^ac  +  ice  -f-  ^ab  half  the  sura  of  all  the  three 
sides  of  the  triangle  abc,  or  ck  =  ^s,  calling  s  the  sum  of  those  three 
.sides.  Again  hk  =  hi=  iiF=^AB,or  ki.=»  ab;  theref.  cl  =5  ck — 
KL  =»s  Js  —  AB     and  AK   =^  CK  — CA  «•  ^s  —  AC,  and  al  =a  dk  =» 

ck  — CD  =   |s-CB. 

Now,  by  the  first  riile.  AG  .  co  =  the  A  ACE,  and  AG  .  to  =  th® 
A  ABE.  theref  AG  .  cf  —  /^  ACB*  Also  by  the  parallels,  ag  : 
CO  : :  DF  or  lA  :  cf,  theref.  ag  cf=(  A  acb=)  cg  .lA  i=  cg.  df^ 
conseq.  ag    cf  .  co  .  df  ==»  /^sacb. 

But  CO  .  CF  i=  CK  .  ci'  =^s.  ^s— AB,  and  ag  .  DF  =  AK  .  AL 
=  ii  —  AC  ^s  —  BC  ;  theref,  ag  .  cF  •  CG?  •  df  =s=  /^2  acb  =  ^s. 


^s  —  AB    \s  —  AC.  is  —  BC  is  the  square  of  the  area  of  the  triangle 
ABC      q^E.  D. 

Other'oiise- 
Because  the  rectangle  ag  .  cf  =  the  /^  abc,  and  since  Co  : 
AG  :  :  CF  :  DF  drawing  the  first  and  second  terms  into  CF, 
and  the  third  and  fourth  into  Ao,  the  propor.  becomes  co  .  cf  : 
AG  cf::  ag-gf:  AG.  Dp.orco.cF:  A  abc  : :  A  abc  :  bg  .  df 
that  is,   the  ^  abc  is  a  mean  proportional  between  co  •  CF  and 

AG  .  DF,  OP  between  |s.  As  —  ab  and  |s  —  AC'  ^s  —  BC,        <i:  e*  d« 

Ex.  1. 


4t6^  MENSURATION 

Ex.  1.  To  find  the  area  of  the  triangle  whose  three  side* 
are  20,  30,  40. 

20  45                      45  45 
30  20                     30  40 
40  —                     —.  — 
26  1st  rem       15  2d  rem.  5  3d  rem. 


45  half  sum 

Then  45  X  25  X  15  X  5  =  84375, 
The  root  of  ^  which  is  290-4737,  the  area. 

Ex.  2.  How  many  square  yards  of  plastering  are  in  a  tri- 
angle, whose  sides  are  30,  40,  60,  feet  ?  Ans.  66f . 

Ex.  3.  How  many  acres,  &c.  contains  the  triangle,  whose 
sides  are  2569,  4900,  5025  links  ? 

^  Ans.  61  acres,  1  rood,  39  perches. 


PROBLEM  m. 


To  find  the  Area  of  a  Trapezoid, 

Add  together  the  two  parallel  sides  ;  then  multiply  thetf 
sum  by  the  perpendicular  breadth,  or  the  distance  betwcett 
them  ;  and  take  half  the  product  for  the  area.  By  Geom. 
theor.  29.. 

Ex.   1.   In  a  trapezoid,  the  parallel  sides  are  750  and  1225, 
and  the  perpendicular  distance  between  them  1540  links  :  to 
find  the  area. 
1225 
750 

1975  X  770  =  152075  square  hnks  =  15  acr.  33  perc, 

Ex.  2.  How  many  square  feet  are  contained  in  the  plank, 
whose  length  is  12  ieet  6  inches,  the  breadth  at  the  greater 
end  15  inches,  and  at  the  less  end  11  inches  ? 

Ans.  13if  feet. 

Ex.  3.  In  measuring  along  one  side  ab  of  a  quadrangular 
field,  that  side,  and  the  two  perpendiculars  let  fall  on  it  from 
the  two  opposite  corners,  measured  as  below,  required  the 
content. 


OF  PLANES. 

kf  =     no  links 

A^  =     746 

AB    =    1110 

CP   =     352 

Dft  =     595 

Ans.  4  acres,  1  rood,  6-792  perches. 


AP 


407 


(I      B 


PROBLEM  IV. 


To  find  the  Area  of  any  Trapezium. 

Divide  the  trapezium  into  t'\'0  triangles  by  a  diagonal  ; 
then  find  the  areas  of  these  triangles,  and  add  them  together. 

Or  thus,  let  fall  two  perpendiculars  on  the  diagonal  from 
the  other  two  opposite  angles  ;  then  add  these  two  perpen- 
dirulars  together,  and  multiply  that  sum  by  the  diagonal,  tak- 
ing half  the  product  for  the  area  of  the  trapezium. 

Ex*  1.  To  6nd  the  area  of  the  trapezium,  whose  diagonal 
is  42,  and  the  two  perpendiculars  on  it  16  and  18. 
Here  16  +  18  =    34,  its  half  is  17. 
Then  42  X  11  =  114  the  area. 

Ex.  2.  How  many  square  yards  of  paving  are  in  the  tra- 
pezium, whose  diagonal  is  65  feet,  and  the  two  perpendiculars 


let  fall  on  it  28  and  33i  feet  ? 


Ans.  222  -JJ  yards. 


Ex.  3.  In  the  quadrangular  field  abcd,  on  account  of  ob- 
structions there  could  only  be  taken  the  following  measures, 
viz.  the  two  sides  b€  i^66  and  ad  ^^20  yards,  the  diagonal  ag 
378,  and  the  two  distances  of  the  perpendiculars  from. the 
ends  of  the  diagonal,  namely,  ae  100,  and  cf  70  yards,- 
Re':|uiredthe  construction  of  the  figure,  and  the  area  in  acres, 
when  4840  square  yards  make  an  acre  ? 

Ans.  17  acres,  2  roods,  21  perches. 

PROBLEM  V^. 

To  find  the  Area  of  an  Irregular  Polygon, 

Draw  diagonals  dividing  the  proposed  polygon  into  trape- 
ziunis  and  triangles.  Then  find  the  areas  of  of  all  these  sepa- 
rately and  add  them  together  for  the  content  of  the  whole 
polygon. 

EXAMP1*E. 


408  MENSURATION^ 

Example.  To  find  the  content  of  the  irregular  figure 
ABCDEFGA,  in  which  are  given  the  following  diagonals  and  per- 
pendiculars :  namely, 


A- 


AC  56 
FD  62 
cc  44 
cm  13 
Bn  18 
60  12 
Ep  8 
Dq  23 

Ans.  1878J^ 


PROBLEM  VI. 

Tojlnd  the  Area  of  a  Regular  Polygon. 

Rule  I.  Multiply  the  perimeter  of  the  polygon,  or  sum 
«f  its  sides,  by  the  perpendicular  drawn  from  its  centre  on 
•ne  of  its  sides,  and  take  half  the  product  for  the  area*. 

Ex.  I.  To  find  the  area  of  the  regular  pentagon,  each  side 
being  26  feet,  and  the  perpendicular  from  the  centre  oa 
each  side  is  17-2047737. 

Here' 26  X  6  ==  125  is  the  perimeter. 
And  17-2047737  X  126  =  2150"6967126. 
Its  half  1076'29835625  is  the  area  sought. 

Rule  II.  Square  the  side  of  the  polygon  ;  then  multiply 
that  square  by  the'  tabular  area,  or  multiplier  set  against  its 
name  in  the  following  table,  and  the  product  will  be  the 
areat. 

No. 


*  This  is  only  in  effect  resolving  the  polygon  into  as  many  equal 
triangles  as  it  has  sides,  by  drawing  lines  irom  the  centre  to  all  the 
angles  ;  then  finding  their  aicas,  and  adding  them  ail  together. 

f  This  rule  is  founded  on  the  property,  that  like  polygons,  being 
similar  figures,  are  to  one  another  as  the  equates  of  tlieir  like  sides  ; 
•which  is  proved  in  the  Geom.  theor.  89.  Now,  the  multipliers  in  the 
table,  are  the  areas  of  the  respective  polygons  to  the  side  1.  W  hence 
iha  rule  is  manifest.  . 


OF  PLANES. 


405 


Sides. 

Names. 

Areas,  or 
Multipliers. 

3 

Trigon  or  triangle 

0-4330^27 

4 

Tetragon  or  square 

1-0000000 

5 

Pentagon 

1-7204774 

6 

Hexagon 

2-5980762 

7 

Heptagon 

3-6339124 

8 

Octagon 

48284271 

9 

Nonagon 

6-1818242 

10 

Decagon 

7-6942088 

11 

Undecagon 

9-3656399 

12 

Dodecagon 

11-1961524 

Exam.   Taking  here  the  same  example  as  before,  namely, 
a  pentagon,  whose  side  is  25  feet. 
Then  25^  being  =  625, 
And  the  tabular  area  1-7204774  ; 
Theref.  1-7204774  X  625  =  1075-298375,  as  before. 

Ex.  2.  To  find  the  area  of  the  trigon,  or  equilateral  tri- 
angle, whose  side  is  20,  Ans.  173-20508. 

Ex.  3.  To  find  the  area  of  the  hexagon  whose  side  is  20. 

Ans.  1039-23048, 

Ex.  4.  To  find  the  area  of  an  octagon  whose  side  is  20. 

Ans.  1931-37084. 

Ex.  6.  To  find  the  area  of  a  decagon  whose  side  is  20. 

Ans.  3077-68352. 


^ote.  The  areas  in  the  table  to  each  side  1 
thay  be  computed  in  the  f  )Uowing  manner  : 
From  the  centre  c  of  the  polygon  draw  lines 
to  every  angle,  dividing  the  whole  figure  into 
as  many  equal  triangles  as  the  polygon  has 
sides  ;  and  let   ABC  be  one  of  those  triangles, 

the    perpendicular   of    which    is    cd.     Divide  . 

360  degrees  by  the  number  of  sides  in  the  po-  A      D      B 

lygon,  the  quotient  gives  the  angle  at  the  centre  acb.  The  half 
of  this  gives  the  angle  acd  ;  and  Uiis  taken  from  90°.  leaves  the 
IJ:T11  ^*'""  -^^^r^ihe,  as  radius  is  to  ad,  so  is  t.ng.  angle 
fh  h^lf^'holPfP^''^'''"'^!;''^'  '^i?'"  perpendicular,  multiplied  by 
'!«  u  •  \^t^  ^'''  ^'-'"^^  ^^«  *'^*  of  the  triangle  abc  :  which  bein? 
Cl7  ^^?vi7  >''  "''";'"*  "^  '^^  '^'""^'^^'  "^  «f  tne  sides  of  the  ^? 
fifures.^  "^"^^  *'^*'  *'  ^"  '^^  **^^^»  f^^  every  one  of  th« 

^*'"  ^'  53  PROBLEM 


410 


MENSURATION^ 


PROBLEM  VII. 

To  find  the  Diameter  and  Circumference  of  any  Circhy  the  one  \ 
from  the  other. 

This  may  be  done  nearly  by  either  of  the  two  follomD|;^ 
proportions,  ' 


viz.  As  7  is  to  22,  so  is  the  diameter  to  the  circumference#|^ 
Or,  As  1  is  to  3-1416,  so  is  the  diameter  to  the  circumfe*| 

rpnr.f»*- 


Ex.   1.  To  find  the  circumference  of  the  circle  whose  di2L*7i 

meter  is  20.  ;| 

By  the  first  rule,  as  7  ;  22  :  :  20  :  62#,  the  answer.  1 

Ex.  St  J 


*  For,  let  ABCD  be  any  circle,  whose  centre 
is  B,  and  let  AB,  bc  be  any  two  equal  arcs. 
Draw  the  several  chords  as  in  the  figure,  and 
join  BE  ;  also  draw  the  diameter  DA,  which 
produce  to  f,  till  bb  be  equal  to  the  chord  bd. 

Then  the  two  isosceles  triangles  deb,  dbf, 
are  equiangular^  because  they  have  the  angle  at 
D  Cf)mmon  ;  consequently  oe  :  db  :  :  DB  :  df. 
But  the  two  triangles  afb,  dcb  are  identical, 
or  equal  in  all  respects,  because  they  have  the 
angle  f  «  the  angle  bdc,  being  each  equal  to 
the  angle  ADB,  these  being  subtended  by  the 
equal  arcs  AB,  bc  ;  also  the  exterior  angle  FAB  of  the  quadrangle 
ABCD,  is  equal  to  the  opposite  interior  angle  at  c  j  and  the  two  tri- 
angles have  also  the  side  bf  =^  the  side  bd  ;  theref'-'re  the  side  af 
is  also  equal  to  the  side  DC  Hence  the  proportion  above,  viz.  DE  : 
DB  : :  DB  :  DF  ■=«  DA  +  AF,  becomes  de  :  db  : :  db  :  2de  +  do. 
Then,  by  taking  the  rectangles  of  the  extremes  and  means,  it  is  db2 
="  2de2   -|-  de  .  DC. 

Now,  if  the  radius  DE  be  taken  «■  1,  this  expression  becomes 


DB»  -=«  2  +  DC,  and  hence  the  root  db  =>  V2  -I-  nc  That  is.  If 
the  measure  of  the  supplemental  chord  of  any  arc  be  increased  by  the 
number  2,  the  square  root  of  the  sum  will  be  the  supplemental  chord 
of  half  that  arc. 

Now,  to  apply  this  to  the  calculation  of  the  circumference  of 
the  circle,  let  the  arc  AC  be  taken  equal  to  ^  of  the  circumfe- 
rence, and  be  successively  bisected  by  the  above  theorem  :  thus, 
the*  chord  AC  of  ^  of  the  circumference,  is  the  side  of  the  in- 
scribed regular  hexagon,  and  is  therefore  equal  to  the  radius  ae 
or  1 :  hence,  in  the  right-angled  triangle    acd,  it  will  be  dc    -■ 

V'ad»-ac* 


OF  PLANES. 


411 


Ex.  2.  If  the  circumference  of  the  earth  be  26000  miles, 
what  is  its  diameter  ? 

By  the  2d  rule,  as  3-1416  :  1  :  :  26000  :  7967^  nearly 
the  diameter. 

Note  by  R.  Adrain.  Hanag  applied  my  new  theory  of 
most  probable  values  to  the  determination  of  the  magnitude 
and  Sgure  of  the  earth,  I  found  the  true  mean  diameter  of  the 
earth,  taken  as  a  globe,  to  be  7918  7  Enghsh  miles,  and  conse- 
quently its  circumference  24877'4  E.  miles,  and  a  degree  of 
a  great  circle  equal  to  69-1039  miles. 


y/3  =»  1-7320508076,  the  supplemental 


, 

a 

12 

i"s 

^ 

I 

48 

I 

<»    y 

192 

0)  ^ 

-a  42 

7k 

T*T 

fc- 

<B 

Lti'sj  J 

,   ^  AD2  -AC2=s^23  — 16    - 

chord  of  1  of  tlie  |>eriphery. 

Then,  by  the  foregoing  theorem,  by  always  bisectuig  the  arcs, 
and  adding  2  to  the  last  square  root,  there  will  be  found  the  supple- 
metjtai  chords  of  the  I2th,  the  24th,  the  48th,  the  96th,  &c.  parts  of 
the  periphery  ;  thus, 

^3-7320608076  =  1-9318516525^ 

^3'9318516525  =  1-9828897227 

!^3-9828897227  =  1 '9957178465 

^3-9957178465  =  1-9989291743 

^3-9989291743  =  1-9997322757 

^^3-9997322757  =  1-9999330678 

v^3'9999330678  =  1-9999832669 

v/3-9999832669  = 
Since  then  it  is  found  that  3-9i/99832669  is  the  square  of  the  sup- 
plemental chord  of  the  1536th  part  of  the  periphery,  let  this  nuHiber 
be  taken  from  4,  which  is  the  square  of  the  diameter,  and  the  remain- 
der 0'0000167331  will  be  the  square  of  the  chord  of  the  said  1536th 
part  of  the  periphery,  and  consequently  the  root  ^^0-0000167331  = 
0-0040906112  is  the  length  of  that  chord  ;  this  number  then  being 
multiplied  by  1536,  gives  6*2831788  for  the  perimeter  of  a  regular 
polygon  of  1536  sides  inscribed  in  the  circle  j  which^  as  the  sides  of 
the  polygon  nearly  coincide  with  the  circumference  of  the  circle,  must 
also  express  the  length  of  the  circumference  itself,  very  nearly. 

But  now,  to  show  how  near  this  determination 
is  to  the  truth,  let  aq.p  =  0-0040906112  represent 
one  side  of  such  a  regular  poly/;on  of  1586  sides, 
and  SRT  a  side  of  another  similar  polygon  de- 
scribed about  the  circle  ;  and  from  the  centre  e 
let  the  perpendicular  Eq,R  be  drawn,  bisecting  ap 
and  ST  in  q,and  r.  Then  since  aq^s  =  |ap  = 
0-0020453056,  and  EA  =  1,  therefore  eq.»  =  ea^ 
—  Aq3  =s  '9999958167,  and  consequently  its  root 
gives  Eq.  =  -9999979084  ;  then  because  of  the 
parallels  ap,  st,  it  is  eq.:  er  :  :  ap  :  st  :  :  as  the 
whole  inscribed  perimeter  :  to  the  circumscribed  one,  that  is,  as 
■9999979084  :  1  :  :  6  2831788  : 6-2831920  the  perimeter  of  the  circum- 
jBcribed  polygon.    Now,  the  circumference  of  the  circle  being  greater 

than 


412  MENSURATION 

PROBLEM  VIU. 

To  find  the  Length  of  any  Arc  of  a  Circle, 

Multiply  the  decimal  -01745  by  the  degrees  in  the  giveii 
arc,  and  that  product  by  the  radius  of  the  circle,  for  the 
length  of  the  arc*. 

Ex.  1.  To  find  the  length  of  an  arc  of  30  degrees,  the 
radius  being  9  feet.  Ans.  4  7115. 

Ex.  2.  To  find  the  length  of  an  arc  of  12°  10',  or  12«>i, 
the  radius  being  10  feet.  Ans.  2-1231, 

PROBLEM  IX. 

To  find  the  Area  of  a  Circle]. 

Rule  I.  Multiply  half  the  circumference  by  half  the 
diameter.  Or  multiply  the  whole  circumference  by  the 
whole  diameter,  and  take  |  of  the  product. 

Rule 


than  the  perimeter  of  the  inner  polygon,  but  less  than  that  of  the 
outer,  it  must  consequently  be  greater  than  6-2831788, 

but  less  than  6-2831920, 
and  must  therefore  be  nearly  equal  ^  their  sum,  or  6-2831854, 
which  in  fact  is  true  to  the  last  figure,  which  should  be  a  3  instead 
of  the  4. 

Hence,  the  circumference  being  6-2831854  when  the  diameter  is 
2,  it  will  be  the  half  of  that,  or  3-1415927,  when  the  diameter  is  1, 
to  which  the  ratio  in  the  rule,  viz.  1  to  S-14i6  is  very  near.  Also  the 
other  ratio  m  the  rule,  7  to  22  or  1  to  3^  =  3-1428  &,c.  is  another  near 
approximation,    " 

*  it  having  been  f^und,  in  the  demonstration  of  the  foregoing-  prob- 
lem, that  when  the  radius  of  a  circle  is  1,  the  length  of  the  whole  cir- 
Gumftrence  s  6*2831854,  which  consists  of  360  degrees  ;  therefore  as 
3,60*^  :  6  2831854  ::!*>:  01745  &c.  the  length  of  the  arc  of  1  degree. 
Hence  the  decimal  -01745  multiplied  by  any  number  of  degrees,  will 
give  the  length  of  ^he  arc  of  those  degrees.  And  because  the  cir- 
cumferences and  arcs  are  in  proportion  as  the  diameters,  or  as  the 
1-adii  of  the  circles,  therefore  as  the  radius  1  is  to  any  other  radius  r, 
so  is  the  lenffth  of  the  arc  above  mentioned,  to  -01745  x  degrees  in 
the  arc  X  *^t  which  is  the  length  of  that  arc,  as  in  the  rule. 

t  The  first  rule  is  proved  in  the  Geom.  theor.  94. 

And  the  2d  and  3d  rules  are  deduced  f;om  the  first  rule,  in  this 
manner.— By  that  rule,  c^c  •—•  4  is  the  area,  when  d  denotes  the  diame- 
ter. 


OF  PLANES.  '4Vd 

Rule  II.  Square  the  diameter,  and  moltiply  that  square  by 
the  decimal  -7864,  for  the  area. 

Rule  IH.  Square  the  circumference,  and  multiply  that 
square  by  the  decimal   07958. 

Kx.  1.  To  find  the  area  of  a  circle  whose  diameter  is  10, 
and  its  circumference  31-416. 

By  Rule  3. 
31-416 
31-416 


By  Rule  1, 

By  Rule  2. 

31-416 

•7854 

10 

102=      100 

4)314-16  „«  . .  986-965 

78-54  -07958 


78-54 


So  that  the  area  is  78-54  by  all  the  three  rules. 

Bx.  2     To  find  the  area  of  a  circle,  whose  diameter  is  7, 
and  circumference  22.  Ans.  38|. 

Ex.  3.    How  many  square  yards  are  in  a  circle  whose  dia- 
meter is  3i  feet  ?  Ans.  l-d69. 

Ex.  4.  To  find  the  area  of  a  circle  whose  circumference 
is  12  feet.  Ans.  11-4595. 

PROBLEM  X.^ 

To  find  the  Area  of-  a  Circular  Ringy  or  of  the  Space  included 
between  the  Circumferences  of  two  Circles  ;  the  one  being 
contained  within  the  other. 

Take  the  difference  between  the  areas  of  the  two  circles, 
as  found  by  the  last  problem,  for  the  area  of  the  ring. — Or, 


ter,  and  c  the  circumference.  But,  by  prob.  7,  c  is  =*  3-1416t//  there- 
fore the  said  area  dc  -j-  4,  becomes  d  X  3*1416^/  -i-  4  =»  -7854^3, 
which  gives  the  2d  rule. — Also,  by  the  same  prob.  7,  d  \s  «—  c  -r- 
3-1416  ;  therefore  again  the  same  first  area  dc  -f-  4,  becomes  c  4- 
3-1416  X  c  -r  4  =»  c2  -7-  12*5664,  which  is  =-.  c^  x  -07958,  by  taking 
the  reciprocal  of  12-5664,  or  changing  that  divisor  into  the  multiplier 
•07958  ;  which  gives  the  3d  rule. 

Carol.  Hence,  the  areas  of  different  circles  are  in  proportion  to  one 
another,  as  the  square  of  their  diameters,  oras  the  square  of  their 
circumferences  ;  as  before  proved  in  the  Geom.  theor.  93. 

which 


414  MENSURATION 

which  is  the  same  thing,  subtract  the  square  of  the  less  dia- 
meter from  the  square  of  the  greater,  and  multiply  their  diflf- 
erence  by  -7854. — Or  lastly,  multiply  the  sum  of  the  dia- 
meters by  the  difference  of  the  same,  and  that  product  by 
•7854  ;  which  is  still  the  same  thing,  because  the  product  of 
M;he  sum  and  difference  of  any  two  quantities,  is  equal  to  the 
difference  of  their  squares. 

Ex.  1.  The  diameters  of  two  concentric  circles  being  10 
and  6,  required  the  area  of  the  ring  contained  between  their 
circumferences. 

Here  10  -f  6  =  16  the  sum,  and   10  —  6  =  4  the  diff. 

Therefore  -7864  X    16  X  4  =  -7854  X  64  =  60-2656, 

the  area. 

Ex.  2.     What  is  the  area  of  the  ring,  the  diameters  of 

whose  bounding  circles  are  10  and  20  ?  Ans.  235*  62. 


PROBLEM  XI. 

To  find  the  Area  of  the  Sector  of  a  Circle. 

Rule  I.  Multiply  the  radius,  or  half  the  diameter,  by 
half  the  arc  of  the  sector,  for  the  area.  Or,  multiply  the 
whole  diameter  by  the  whole  arc  of  the  sector,  and  take  I 
of  the  product.  The  reason  of  which  is  the  same  as  for  the 
first  rule  to  problem  9,  for  the  whole  circle. 

Rule  II.  Compute  the  area  of  the  whole  circle  :  then  say,^ 
as  360  is  to  the  degrees  in  the  arc  of  the  sector,  so  is  the 
area  of  the  whole  circle  to  the  area  of  the  sector. 

This  is  evident,  because  the  sector  is  proportional  to  the 
length  of  the  arc,  or  to  the  degrees  contained  in  it, 

Ex.  1.  To  find  thie  area  of  a  circular  sector,  whose  an^ 
contains  18  degrees  ;   the  diameter  being  3  feet  ? 

1.  By  the  1st  Rule. 

First,  3- 14 16  X  3  =  9-4248,  the  circumference. 

And  360  :  18  :  ;  9-4248  :  -47124,  the  len2;th  of  the  arc. 

Then  -47124  X  3 -r- 4  =  1-41372  ~  4  =^ -35343,  the  area, 

2.  By  the  2d  Rule. 

First,  -7854  X  3^  =  7-0686,  the  area  of  the  whole  circle. 
Then,  as  3$0   :    18   :  :  7-0686  :  -36343,  the  area  of  the 
sector 

Ex.  2. 


OF  PLANES.  415 

Ex.  2.  To  find  the  area  of  a  sector,  whose  radius  is  10,  and 
arc  20.  Ans.   100. 

Ex.  3.  Required  the  area  of  a  sector,  whoseT  radius  is  35, 
and  its  arc  containing  147°  29*.  Ans.  804  3986 


PROBLEM  Xn. 

To  find  the  Area  of  a  Segment  of  a  Circle. 

Rule  I.  Find  the  area  of  the  sector  having  the  same  arc 
with  the  segment,  by  the  last  problem. 

Find  also  the  area  of  the  triangle,  formed  by  the  chord  of 
the  segment  and  the  two  radii  of  the  sector. 

Then  add  these  two  together  for  the  answer,  when  the 
segment  is  greater  than  a  semicircle  ;  or  subtract  them  when 
it  is  less  than  a  semicircle. — As  is  evident  by  inspection. 

Ex.  1.  To  find  the  area  of  the  segment  acbda,  its  chord  ab 
being  12,  and  the  radius  ae  or  ce  10. 

First,  As  AE  :  sin.    ^d  90**    :  :  ad  :  sin.  C 

36*>  62'i  =  36-87  degrees,  the  degrees  in  the         A/^T^B 
^    AEc   or  arc  ac.       Their  doubie,   73*74,  /    \^.  '' 

are  the  degrees  in  the  whole  arc  acb.  \ 

Now  ••7864   X  400  =  314-16,  the  area  of 

the  whole  circle.  ^ 

Therefore  360«»  :  73-74  : :  314  16  :  64-3504,  area  of  the 
sector  ACBE. 


^gain,  ^  AE3-ADa  =  ^  100-36  =  ^  64  =  B   =   de. 
Theref.  ad    X  de  =  6  X  8  =  48,  the  area  of  the  trian- 
gle AEB. 
Hence  sector  acbe —  triangle  aeb  =  16-3504,  area  of  seg. 


Rule  II.  Divide  the  height  of  the  segment  by  the  diameter, 
and  find  the  quotient  in  the  column  of  heights  ia  the  following 
tablet  :  Take  out  the  corresponding  area  in  the  iv^xi  cohiniu 
on  the  right  hand  ;  and  multiply  it  by  the  square  of  the  cir- 
cle's diameter,  for  the  area  of  the  segment*. 

Note. 


*  The  truth  of  this  rule  depends  on  tlie  principle  of  r-imllar 
plane  figures,  which  are  to  one  another  as  tlie  square  of  (.heh* 
like  linear  dimensions.    The  segments  in  the  table  are  thost  nf  a 

circle 


41S 


MENSURATION 


Note.  When  the  quotient  is  not  found  exactly  in  the  table, 
proportion  may  be  made  between  the  next  less  and  greater 
area,  in  the  same  manner  as  is  done  for  logarithms,  op  any 
other  table. 


Table  of  the  Areas  of  Circular  Segments. 


^1 

0)  CCl 

% 

0) 

o    ^ 

as 

<% 

ffi 

^    0) 

<3^  / 

IX 

•11990 

X 
•31 

-^1 

•'^0738 

•41 

^     0) 

-303  r9 

•01 

•00133 

•T7 

•04701 

.21 

•02 

•00376 

•12 

•06339 

•22 

•12811 

•32 

•21667 

•42 

•31304 

•03 

•00687 

■13 

•06000 

•23 

13646 

•33 

•22603 

43 

•32293 

•04 

01064 

•14 

•06683 

•24 

14494 

•34 

•23647 

•44 

•33284 

•05 

•01468 

•16 

•07387 

•26 

•16354 

•36 

•24498 

•45 

•34278 

•06 

•01924 

•16 

•08111 

26 

•16226 

•3fi 

•2545f> 

•46 

•36274 

•07 

•02417 

17 

•08863 

•27 

•17109 

•37 

•26418 

47 

•36272 

•08 

•02944 

-18 

•09613 

28 

•18002 

38 

•2738(> 

•48 

•3727D 

•09 

•03602 

•19 

10:^90 

•29 

18906 

•39 

•2835^; 

•49 

•38270 

•10 

•04088 

•20 

•11182 

•30 

•19817 

•40  •29337) 

•50 

•39270 

Ex.  2.  Taking  the  same  example  as  before,  in  which- are 
given  the  chord  ab  12,  and  the  radius  10,  or  diameter  20. 

And  having  found,  as  above,  de  =  8  ;  then  ce  —  de  =  ci> 
=:  10  —  8  =  2.  Hence,  by  the  rule,  cd  -r-  of  =  2  -f-  20  =ar 
1  the  tabular  height.  This  being  found  in  the  first  column  of 
of  the  table,  the  corresponding  tabular  area  is  •04088.  Then 
•04O88  X  202  =  -04088  X  400  =  16-362,  the  area,  nearly 
the  same  as  before. 

Ex.  3.  What  is  the  area  of  the  segment,  whose  height  is  18^ 
and  diameter  of  the  circle  60  ?  Ans.  636-375. 

Ex.  4.  Required  the  area  of  the  segment  whose  chord  is  16, 
the  diameter  being  20  ?  Ans.  44*728. 


circle  -whose  diameter  Is  1  ;  and  the  first  column  contains  the  cor- 
responding heights  or  versed  sines  divided  by  the  diameter.  Thus 
then,  the  area  of  the  similar  segment,  taken  from  the  table,  and  mul- 
tiplied by  the  square  of  the  diameter,  gives  the  area  of  the  segment  t» 
this  diameter. 


PROBLEM 


OF  PLANES. 


4J7 


PROBLEM  Xia 


To  measure  long  Irregular  Figures, 

Take  or  measure  the  brea<lth  at  both  ends,  and  at  se- 
veral places  at  equal  distances.  Then  add  together  all  these 
intermediate  breadths  and  half  the  two  extremes,  which 
sum  multiply  by  the  length,  and  divide  by  the  number  of 
parts  for  the  area*. 

JSToie,  If  the  perpendiculars  or  breadths  be  not  at  equal 
distances,  compute  all  the  parts  separately,  as  so  many  tra- 
pezoids, and  add  them  all  together  for  the  whole  area. 

Or  else,  add  all  the  perpendicular  breadths  together,  and 
divide  their  sum  by  the  number  of  them  for  the  mean  breadth, 
to  multiply  by  the  length  ;  which  will  give  the  whole  area, 
not  far  from  the  truth. 

Ex.  1.  The  breadths  of  an  irregular  figure,  at  five  equi- 
distant places,  being  8-2,  7-4,  9-2,  10-2,  86  ;  and  the  whole 
length  39  ;  required  the  area  ? 

8*2  35'2  sum, 

8-6  39 


2)  16-8  sum  of  the  extremes. 

8-4  mean  of  the  extremes. 

7-4 
9-2 
10-2 

35-2  sum. 


3168 
105b 

4)  1372-8 


343-2  the  area. 


Ex. 


♦  This  rule  is  made  out  as  follows  : 
—Let  ABCD  be  the  irregular  piece  , 
having-  ihe  several  breadths  ad  EF  gh, 
IK,  Bc,  at  tlie  equal  distances  ae,  eg, 
oi,  IB.  Let  the  several  breadths  ui  or- 
4ep  be  denoted  by  the  cori-esponding  let- 
ters a,  b,  c,  d,  e,  and  the  whole  length 

ab  by  /  ;  then  compute  the  areas  of  he  parts  into  which  the  figure  is 
divided  by  the  perpendiculars,  as  so  many  trapezo  ds,  by  prob.  3,  and 
add  them  all  together     Thus,  the  sum  of  the  parts  is, 
a-^  b  b-\'  c  c  +  d  d  +e 

' X    AE  -1 X  EG  4-  X    GI    -f-  X   IB 

2  2  2  2 

a  ■\.  b  A-fc  c  +  d  d  ■{•  e 

« X  i/  +  X  i/  +  X  i/  -f.  X  H 

2  2  2  2 


«=-  (i«  +  6  +  c  -h  rf  +  le)  X  i/  -=  (m  +  A  +  c  +  </)  J/ 


Vol.  I. 


wbicJi 


i4 


U8  MENSURATION 

Ex.  2.  The  len^h  of  an  irregular  figure  being  84,  and  the 
breadths  at  six  equidistant  places  17*4,  20*6,  14-2,  16*5,  20- 1, 
24-4  J  what  is  the  area  ?  Ans.  1560-64. " 

PROBLEM  XIV. 

To  find  the  Area  of  an  Ellisis  or  Oval. 

Multiply  the  longest  diameter,  or  axis,  by  the  shortest  j 
then  multiply  the  product  by  the  decimal  7854,  for  the  area. 
As  appears  from  cor.  2,  theor.  3,  of  the  Ellipse,  in  the  Conic 
Sections. 

Ex.  1.  Required  the  area  of  an  ellipse  whose  two  a\es 
are  70  and  60  Ans.  2746-9. 

Ex.  2.  To  find  the  area  of  the  oval  whose  two  axes  are 
24  and  18.  Ans.  339-2928. 

PROBLEM  XV. 

Tpfind  the  Area  of  any  Elliptic  Segment. 

Find  the  area  of  a  corresponding  circular  segment,  having 
the  same  height  and  the  same  vertical  axis  or  diameter.  Thea 
say,  as  the  said  vertical  axis  is  to  the  other  axis,  parallel  to 
the  segment's  base,  so  is  the  area  of  the  circular  segment 
before  found,  to  the  area  of  the  elliptic  segment  sought. 
This  rule  also  comes  from  cor.  2,  t|»eor.  3  of  the  Ellipse. 

Otherwise  thus.  Divide  the  height  of  the  segment  by  the 
vertical  axis  of  the  ellipse  ;  and  find,  in  the  table  of  circular 
segments  to  prob.  12,  the  circular  segment  having  the  above 
quotient  for  its  versed  sine  :  then  multiply  all  together,  this 
segment  and  the  two  axes  of  the  ellipse,  for  the  area. 

Ex.  1.  To  find  the  area  of  the  elliptic  segment,  whose 
height  is  20,  the  vertical  axis  being  70,  and  the  parallel 
axis  50. 


which  is  the  whole  area,  agreeing*  with  the  rule  :  m  being  the  arith- 
inetical  mean  between  the  txtremes,  or  half  the  sum  of  them  both, 
and  4  the  number  of  the  patts.  And  the  same  for  any  other  number 
of  paits  whatever. 

Here 


OF  SOLIDS.  419 

Here  20  -f-   70  gires  -284  the  quotient  or  versed  sine  ;  to 
which  io  the  table  answers  the  seg.  '18518 
then         70 


12-9tt260 
60 


648-13000  the  area. 

Ex.  2.  Required  the  area  of  an  elliptic  segment,  cut  otf 
parallel  to  the  shorter  axis  ;  the  height  being  10,  and  the 
two  axes  25  and  35.  Ans.  162  03. 

Ex.  3.  To  find  the  area  of  the  elliptic  segment,  cut  oflf 
parallel  to  the  longer  axis  ;  the  'height  being  6,  and  the  axes 
25  and  35.  Ans.  97-8425. 

PROBLEM  XVI. 

To  find  the  Area  of  a  Parabola,  or  its  Segment, 

Multiply  the  base  by  the  perpendicular  height ;  then  take 
two-thirds  of  the  product  for  the  area.  As  is  proved  in  theo- 
rem 17  of  the  Parabola,  in  the  Conic  Sections. 

Ex.  1 .  To  find  the  area  of  a  parabola  ;  the  height  being  2, 
and  the  base  12. 

Here  2  X  12  =  24.    Then  |  of  24  =  16,  is  the  area. 

Ex.  2.  Required  the  area  of  the  parabola,  whose  height 
is  10,  and  its  base  16.  Ans.  106§, 


MENSURATION  OF  SOLIDS. 

BY  the  Mensuration  of  Solids  are  determined  the  spaces 
included  by  contiguous  surfaces  ;  and  the  sum  of  the  measures 
of  these  including  surfaces,  is  the  whole  surface  or  superficies 
of  the  body. 

The  measure  of  a  solid,  is  called  its.  solidity,  capacity,  or 
content. 

Solids  are  measured  by  cubes,  whose  sides  are  inches,  or 
feet,  or  yards,  &c.  And  hence  the  solidity  of  a  body  is  said 
to  be  so  many  cubic  inches,  feet,  yards,  &c.  as  will  fill  its  ca- 
pacity or  space,  or  another  of  an  equal  magoitude. 

The 


420  MENSURATrON 

The  least  solid  measure  is  the  cubic  inch,  other  cubes  being 
laken  from  it  according  to  the  proportion  in  the  following 
table,  which  is  formed  by  cubing  the  linear  proportions. 


Table  of  Cubic  or  Solid  Measures. 

1728  cubic  inches  make  1  cubic  foot 

27  cubic  feet  -  1  cubic  yard 

166f  cubic  yards       -  1  cubic  pole 

64000  cubic  poles       -  1  cubic  furlong 

512  cubic  furlongs  -  1  cubic  mile. 

PROBLEM  I. 

To  Jind  the  Superficies  of  a  Prism  or  Cylinder. 

Multiply  the  perimeter  of  one  end  of  the  prism  by  the 
length  or  height  of  the  solid,  and  the  product  will  be  the  sur- 
face of  all  its  sides.  To  which  add  also  the  area  of  the 
two  ends  of  the  prism,  when  required*. 

Or,  compute  the  areas  of  all  the  sides  and  ends  separately, 
and  add  them  all  together. 

Ex.  1.  To  find  the  surface  of  a  cube,  the  length  of  each 
side  being  20  feet.  Ans.  2400  feet. 

Ex.  2.  To  find  the  whole  surface  of  a  triangular  prism, 
whose  length  is  20  feet,  and  each  side  of  its  end  or  base  18 
inches.  Ans.  91-948  feet. 

Ex.  3.  To  find  the  convex  surface  of  a  round  prism,  or 
cylinder,  whose  length  is  20  feet,  and  the  diameter  of  its  base 
is  2  feet.  Ans.  126-664. 

Ex.  4.  What  must  be  paid  for  lining  a  rectangular  cistern 
with  lead,  at  2d.  a  pound  weight,  the  thickness  of  the  lead 
being  such  as  to  weigh  7lb.  for  each  square  foot  of  surface  ; 
the  inside  dimensions  of  the  cistern  being  as  follow,  viz.  the 
length  3  feet  2  inches,  the  breadth  2  feet  8  inches,  and  depth 
2  feet  6  inches  ?  Ans.  2/.  3s.  lOirf. 


*  The  truth  of  this  will  easily  appear,  by  considering  that  the  sides 
of  any  prism  are  parallelograms,  whose  common  length  is  the  same 
as  the  lenpjth  of  the  solid,  and  their  breadths  taken  all  together  make 
up  the  perimeter  of  the  ends  of  the  same. 

And  the  rulo  is  evidently  the  same  for  the  surface  of  a  cylinder. 

PROBLEM 


OF  SOLIDS.  421 

PROBLEM  IL 

To  find  the  Surface  of  a  regular  Pyramid  or  Cone, 

Multiply  the  perimeter  of  the  base  by  the  slant  height,  or 
perpendicular  from  the  vertex  on  a  side  of  the  base,  and  half 
the  product  will  evidently  be  the  surface  of  the  sides,  or  the 
sum  of  the  areas  of  all  the  triangles  which  form  it.  To  which 
add  the  area  of  the  end  or  base,  if  requisite. 

Ex.  1.  What  is  the  inclined  surface  of  a  triangular  pyra- 
mid, the  slant  height  being  20  feet,  and  each  side  of  the  base 
3  feet?  Ans.  90feet. 

Ex.  2.  Required  the  convex  surface  of  a  cone,  or  circular 
pyramid,  the  slant  height  being  50  feet,  and  the  diameter  of  its 
base  ^  ft^i,  Ans.  667-39. 

PROBLEM  m. 

To  find  the  Surface  of  the  Frustum  of  a  regular  Pyramid  or 
Cone  ;  being  the  lower  part  when  the  top  is  cut  off  hy  a  plane 
parallel  to  the  base. 

Add  together  the  perimeters  of  the  two  ends,  and  multiply 
their  sum  by  the  slant  height,  taking  half  the  product  for  the 
answer  —As  is  evident,  because  the  sides  of  the  solid  are  tra- 
pezoids, having  the  opposite  sides  parallel. 

Ex.  1.  How  many  square  feet  are  in  the  surface  of  the 
frustum  of  a  square  pyramid,  whose  slant  height  is  10  feet  ; 
also,  each  side  of  the  base  or  greater  end  being  3  feet  4  inches, 
and  each  side  of  the  less  end  2  feet  2  inches  ?     Ans.  110  feet. 

Ex.  2.  To  find  the  convex  surface  of  the  frustum  of  a  cone, 
the  slant  height  of  the  frustum  being  12^  feet,  and  the  cir- 
cumferences of  the  two  ends  6  and  8*4  feet.         Ans.  90  feet. 

PROBLEM  IV. 

To  find  the  Solid  Content  of  any  Prism  or  Cylinder. 

Find  the  area  of  the  base,  or  end,  whatever  the  figure 
of  it  may  be  ;  and  multiply  it  by  the  length  of  the  prism  or 
cylinder,  for  the  solid  content*. 


*  This  rule  appears  from  the  Geom.  theor.  110,  cor.  2-    The  same 
ift  more  particularly  shown  as  follows  ;  Let  the  annexed  rectangular 

parallelopipedon 


422 


MENTSURATION 


Note.  For  a  cube,  take  the  cube  of  its  side  by  multiplying 
this  twice  by  itself ;  and  for  a  paraJlelopipedon,  multiply  the 
length,  breadth  and  depth  all  together,  for  the  content. 

Ex.  1.  To  find  the  solid  content  of  a  cube,  whose  side  is  24 
inches.  Ans.  13824. 

Ex.  2.  How  many  cubic  feet  are  in  a  block  of  marble,  it* 
length  being  3  feet  2  inches,  breadth  2  feet  8  inches,  and 
thickness  2  feet  6  inches  ?  Ans.  21|. 

Ex.  3.  How  many  gallons  of  water  will  the  cistern  con- 
tain, whose  dimensions  are  the  same  as  in  the  last  example, 
when  282  cubic  inches  are  contained  in  one  gallon  ? 

Ans.  129if 

Ex.  4.  Required  the  solidity  of  a  triangular  prism,  whose 
length  is  10  feet,  and  the  three  sides  of  its  triangular  end  or 
base  are  3.  4.  6.  feet.  Ans.  60. 

Ex.  5.  Required  the  content  of  a  round  pillar,  or  cylinder, 
whose  length  is  20  feet,  and  circumference  6  feet  6  inches. 

Ans.  48-1459  feet. 


% 


parallelopipedon  be  the  solid  to  be  mea- 
sured, and  the  cube  p  the  solid  measu- 
ring unit,  its  side  being  1  inch,  or  1  foot, 
&c.  i  also,  let  the  length  and  breadth  of 
the  base,  abcd  and  also  the  height  au, 
be  each  divided  into  spaces  equal  to  the 
length  of  the  base  of  the  cube  p  name- 
ly/here  3  in  the  length  and  2  in  the 
breadth,  making  3  times  3  or  6  squares 
in  the  base  AC,  each  equal  to  the  base  of 
the  cube  p.  Hence  it  is  manifest  that 
the    parallelopipedon   will   contain  the  -"*  -O 

cube  p,  as  many  times  as  the  base  ac  contains  the  base  of  the  cube, 
repeated  as  often  as  the  height  AH  contains  the  height  of  the  cube. 
That  is,  the  content  of  any  parallelopipedon  is  found,  by  muLiplymg 
the  area  of  <he  base  by  the  altitude  of  that  solid. 

And,  because  all  prisms  and  cylinders  are  equal  to  parallelopipe- 
dons  of  equal  bases  and  altitudes>  by  Geom-  theor.  108,  it  follows  that 
the  rule  is  general  for  all  such  solids,  whatever  the  figure  of  the  base 
may  be. 


F 


FEOBUi^ 


OF  SOLIDS,  4^3 

PROBLEM  V. 

Ihjlfid  the  Content  of  any  Pyramid  or  Cone, 

Find  the  area  of  the  base,  and  multiply  that  area  by  the 
perpendicular  height ;  then  take  ^  of  the  product  for  the 
content*. 

Ex.  1.  Required  the  solidity  of  the  square  pyramid,  each 
side  of  its  base  being  30,  and  its  perpendicular  height  26. 

Ans.  75G0. 

Ex.  2.  To  find  the  content  of  a  triangular  pyramid,  whose 
perpendicular  height  is  30,  and  each  side  of  the  base  3. 

Ans.  38-97117. 

Ex.  3.  '  To  find  the  content  of  a  triang^ilar  pyramid,  its 
height  being  14  feet  6  inches,  and  the  three  sides  of  its  base 
5,  6,  7  feet.  Ahs.  71-0352. 

Ex.  4.  What  is  the  content  of  a  pentagonal  pyramid,  its 
height  being  12  feet,  and  each  side  of  its  base  2  feet  ? 

Ans.  27-5276. 

Ex.  5.  What  is  the  content  of  the  hexagonal  pyramid, 
whose  height  is  6*4  feet,  and  each  side  of  its  base  6  inches  ? 

Ans.  1  38664  feet. 

Ex.  6.  Required  the  content  of  a  cone,  its  height  being 
10^  feet  and  the  eircumference  of  its  base  9  feet. 

Ans.  22-56093. 

PROBLEM  VI, 

To  find  the  Solidity  of  the  Frustum  of  a  Cone  or  Pyramid, 

Add  into  one  sum,  the  areas  of  the  two  ends,  and  the 
mean  proportional  between  them  ;  and  take  ^  of  that  sum 
for  a  mean  area  ;  which  being  multiplied  by  the  perpendicu- 
lar height  or  length  of  the  frustum  will  give  its  contentt. 

fiote. 


*  Thisxule  follows  from  that  of  the  prism,  because  any  pyramid  is 
§■  of  a  prism  of  equal  base  and  altitude  j  by  Geom.  theor  1 15,  cor.  1 
and  2. 

f  Let  ABCD  be  any  pyramid,  of  whxh  bcdgfe  is  a  frustum. 
And  pat  a^  for  the  area  of  th^  base  BCD,  b^  the  area  of  the  top 


424  MENSURATION 

Hote.  This  general  rule  may  be  otherwise  expressed,  as 
follows,  when  the  ends  of  the  frustum  are  circles  or  regular 
polygons.  In  this  latter  case,  square  one  side  of  each  poly- 
gon, and  also  multiply  the  one  side  by  the  other  ;  add  all 
these  three  products  together  ;  then  multiply  their  sum  by  the 
the  tabular  area  proper  to  the  polygon,  and  take  one-third  of 
the  product  for  the  mean  area  to  be  multiplied  by  the  length, 
to  give  the  solid  content.  And  in  the  case  of  the  frustum 
of  a  cone,  the  ends  being  circles,  square  the  diameter  or  the 
circumference  of  each  end,  and  also  multiply  the  same  two 
dimensions  together  ;  then  take  the  sum  of  the  three  products 
and  multiply  it  by  the  proper  tabular  number,  viz.  by  '7864 
when  the  diameters  are  used,  or  by  07958  in  using  the  cir- 
cumferences ;  then  taking  one- third  of  the  product  to  multiply 
by  the  length,  for  the  content. 

Ex.  1.  To  find  the  number  of  solid  feet  in  a  piece  of  timber, 
whose  bases  are  squares,  each  side  of  the  greater  end  being 
16  inches,  and  each  side  of  the  less  end  6  inches  ;  also,  the 
length  or  perpendicular  altitude  24  feet.  Ans.   19^. 

Ex.  2.  Required  the  content  of  a  pentagonal  frustum,  whose 
height  is  6  feet  each  side  of  the  base  18  inches,  and  each  side 
of  the  top  or  less  end  6  inches.  Ans.  931 925  feet. 


EFG,  A  the  height  ih  of  the  frustum,  and  c  the 
height  A I  of  the  top  part  above  it.  Then 
c  -f  A  =  AH  is  the  height  of  the  whole  pyra- 
mid. 

Hence,  by  the  last  prob.  i  as  (c  -f  A)  is  the 
content  of  the  whole  pyramid  abcd,  and  ^b^c 
the  content  of  the  top  part  AEFg  ;  therefore 
the  difference  laS  (c  4-  h)—^b'^  c  is  the  content 
of  the  frustum  bcdgfe.  But  the  quantity  c 
being  no  dimension  of  the  frustum,  it  must  be  ex- 
pelled from  this  formula,  by  substituting  its  value,  found  in  the 
following  manner.  By  Geom.  theor.  IIJ,  a^  :  62  : :  (c  -|-  A)3  :  c2,  or 
d  :  ^  :  :  c  +  A  :  c,  hence  (Geom.  th.  69)  a  —  6  :  6  :  :  h  :   e,    and 

bh  ah 

a —6  :  a\:b\  c+A  ;  hence  therefore  c  <=3 and  c+A  «=  — — —  ; 

a  — 6  a-—b 

then    these    values  of  c    and   c   -f-    A   being   substituted  for  them 
in  the   expression  for  the  content  of  the  frustum,  gives    that   con- 
ah  bh  a3  ^bz 

tent  =  1  a2  X ^^2  x =  ^AnX  —  ='AA  x  (^^ 

a—  b  a  —  b  a^^b 

4-  «3  +  62  )  ;  which  is  the  rule  above  given  j  ab  being  the  mean  be- 
tween «2  and  ^2 : 

Ex,  6. 


SOLIDS. 


42^ 


Ex.  3.  To  find  thp  content  of  a  conic  frustuna,  the  altitude 
being  18,  the  greatest  JiameLer  8,  and  trie  least  dia:neter  4. 

Ans.  627-7888, 

Ex.  4.  What  is  the  solidity  of  the  frustum  of  a  cone,  the  al- 
titude being  25,  also  the  circumference  at  the  greater  end  be- 
ing 20,  and  at  the  less  end  10  ?  Aos.  464-216. 

Ex.  5.  If  a  cask,  which  is  two  equal  conic  frustums  joined 
together  at  the  bases,  have  its  j^ung  diameter  28  inches,  the 
head  diameter  20  inches,  and  length  40  incoes  ;  how  many 
gallons  of  wine  will  it  hold.  Ans.  79*0613, 

PROBLEM  VII. 


To  find  the  Surface  of  a  Sphere  ^  or  any  Segment, 

Rule  1.     Multiply  the  circumference  of  the  sphere  by 
its   diameter,  and  the  product  will  be  the  whole  surface  oJt 

it*. 

Rule  II, 


*  These  rules  come  from  the  following  theorems  for  the  sur- 
fece  of  a  sphere,  viz.  That  the  said  surfsce  is  equal  to  the  cu)^^ 
surface  of  its  circumscribing  cylinder;  or  that  it  is  equal  to  4  great 
circ  es  of  the  same  sphere,  or  of  the  same  diameter  :  which  are  thuij 
proved. 

Let  A  BCD  be  a  cylinder,  circumscribing 
tJie  sphere  ErcHi  tlie  former  generated 
by  the  rotation  of  the  rectangle  FBClf 
abo'it  the  axi.<i  or  d'ameter  fh  ;  and  the 
latter  by  the  rotation  of  the  seroic-rcle 
■pca  aboat  the  same  diamete:-  fh.  Draw 
tw>  lines  kl,  mx,  perp-ndiciilar  to  the 
axis  intercepting  the  parts  ln,  op,  of  the 
cylinder  and  sphere  ;  then  will  the  ring 
or  cylindric  surface  g-eneruted  by  the  ro- 
taiion  of  LM,  be  <qual  to  the  nng  or  spherical  surface  generated 
by  the  arc  op.  For  first,  suppose  the  parallels  kt.  and  m.v  to  be  in- 
definitely near  together;  drawing"  lo,  and  also  oq.  paraUel  to  ln. 
Then  the  two  triane-les  iko,  oq.p,  being  eqtiiansf-lar,  it  is,  as  op  ; 
o<j.  )r  ln:  ;  lo  or  kl  :  ko  :;  ciscumf  ren .v:>  described  by  kl:  cir- 
cunf.  lescribed  by  ko  ,  theTef>re  the  rect^ns^le  op  X  circumf  of  ko 
is  equal  to  the  rectangle  Ln  X  circumf.  of  kl  ;  that  is,  the  r.nc;'  des- 
cribed by  OP  on  tlie  sphere*  is  equal  to  the  ring  described  by  l.n  on 
the  cyMnder. 

Yot.  I.  56  Am€ 


426  MENSURATION 

Rule  II.  Square  the  diameter  and  multiply  that  square  by 
3«1416,  for  the  surface. 

Rule  III.  Square  the  circumference  ;  then  either  multi- 
ply that  square  by  the  decimal  -3183,  or  divide  it  by  3-1416, 
for  the  surface. 

Note.  For  the  surface  of  a  segment  or  frustum,  multiply  the 
whole  circumference  of  the  sphere  by  the  height  of  the  part 
required. 

Ex.  1.  Required  the  convex  superficies  of  a  sphere,  whos6 
diameter  is  7,  and  circumference  22.  Ans.  154. 

Ex!  2.  Required  the  superficies  of  a  globe,  whose  diameter 
is  24  inches.  Ans.  1809-5616. 

Ex.  3.  Required  the  area  of  the  whole  surface  of  the  earth, 
its  diameter  being  7957|  miles,  and  its  circumference  -25000 
miles.  Ans.  198943750  sq.  miles, 

Ex.  4.  The  axis  of  a  sphere  being  42  inches,  what  is  the 
convex  superficies  of  the  segment  whose  height  is  9  inches  ? 

Ans.  1187-5248  inches. 

Ex.  6.  Required  the  convex  surface  of  a  spherical  zone, 
whose  breadth  or  height  is  2  feet,  and  cut  from  a  sphere  of 
12i- feet  diameter.  Ans.  78-54  feet. 


And  as  this  is  every  where  the  case,  therefore  the  sums  of  any 
corresponding  number  of  these  are  also  equal  j  that  is,  the  whole 
surface  of  the  sphere,  described  by  the  whole  semicircle  fgh,  is 
equal  to  the  whole  curve  surface  6f  the  cylinder,  described  by 
the  height  BC  ;  as  well  as  the  surface  of  any  segment  described  by 
Fo,  equal  to  the  surface  of  the  corresponding  segment  described 
byBL. 

CoroU  1.  Hence  the  surface  of  the  sphere  is  equal  to  4  of  its  great 
circles,  or  equal  to  the  circumference  e fgh,  or  of  DC,  multiplied  by 
the  height  bc,  or  by  the  diameter  fh. 

Corol.  2.  Hence  also  the  surface  of  any  such  part  as  a  segment  or 
frustum,  or  zone,  is  equal  to  the  same  circumference  of  the  sphere, 
multiplied  by  the  height  of  the  said  part.  And  consequently  such 
spherical  curve  surfaces  are  to  one  another  in  tl\e  same  proportion  as 
their  altitudes. 

PROBLEM 


OF  SOLIDS. 


427 


PROBLEM  VIIL 

To  find  the  Solidity  of  a  Sphere  or  Globe, 

Rule  I.  Multiply  the  surface  by  the  diameter,  and  take 
I  of  the  product  for  the  content*.  Or,  which  is  the  same  thing, 
multiply  the  square  of  the  diameter,  by  the  circumference,  and 
take  ^  of  the  product. 

RuLB  II.  Take  the  cube  of  the  diameter,  and  multiply  it 
by  the  decimal  '5236,  for  the  content. 

Rule  III.  Cube  the  circumference,  and  multiply  by 
•01688. 

Ex.  1.     To  find  the  content  of  a  sphere  whose  axis  is  12. 

Ans.  904-7808. 

Ex.  2.  To  find  the  solid  content  of  the  globe  of  the  earth 
supposing  its  circumference  to  be  23000  miles. 

Ans.  263858149120  miles. 

PROBLEM  IX. 


To  find  the  Solid  Content  of  a  Spherical  Segment, 
t  Rule  I.      From    3  times  the  diameter  of  the 


sphere 
take 


•  For,  put  d=  the  diameter,  c  —  the  circumference,  and  s  =  the 
surface  of  the  sphere,  or  of  its  circumscribing  cylinder  :  also,  a  = 
the  number  3-1416. 

Then,  i  «  is  =  the  base  of  the  cylinder,  or  one  great  circle  of  the 
sphere  ;  and  d  is  the  height  of  the  cylinder  ;  therefore  ^ds  is  the 
content  of  the  cylinder.  But  |.  of  the  cylinder  is  the  sphere,  by  th.  11 T, 
Geom.  that  is,  |.  of  |cfe,  or  ids  is  the  sphere  ;  which  is  the  first  rule. 

Again,  because  the  surface  »  is  =  ad^  ;  therefore  Xds  =  ^ad^  = 
*'S236d^,  is  the  content,  as  in  the  2d  rule.  Also  d  being  =  c  -r-  a 
therefore  Xad^  "^  ics  -i.  a2  =  -01688^  the  3d  rule  for  the  content, 

I  By  corol.  3,  of  theor.  117*  Geom.it 
appears  that  the  spheric  segment  pfn,  is 
equal  to  the  difference  between  the  cylinder 
ABLo,  and  the  conic  frustum  abmq^. 

But,  putting  rf=  AB  or  TH  the  diameter 
of  the  sphere  or  cyUnder,A=rK  the  height 
of  the  segment,  r  =  pk  the  radius  of  its 
base,  and  a  =  3*1416  ;  then  the  content  of 


the  cone  ABi  is  =:iad*  X  ^  Fi  = 


,ad^'. 


and  by  the  similar  cones  abi,  <i.Mi,  as 


4gS  MENSURATION  OP  SOLIDS. 

take  double  the  height  of  the  segment  :  then  multiply  the 
remainder  by  the  square  of  the  height,  and  the  product  by 
the  decimal  -6236,  for  the  content. 

Rule  II.  To  3  times  the  square  of  the  radius  of  the 
Segment's  base,  add  the  square  of  its  height  ;  then  multiply 
the  sum  by  the  height,  and  the  product  by  'Si^Se,  for  the 
content. 

Ex.  1.  To  find  the  content  of  a  spherical  segment,  of  2 
feet  in  height,  cut  from  a  sphere  of  8  feet  diameter. 

Ans.  41-888. 

Ex.  2.  What  is  the  solidity  of  the  segment  of  a  sphere, 
its  height  being  9,  and  the  diameter  of  its  base  20  1 

Ans.  1795-4244. 


JVofe.  The  general  rules  for  measuring  all  sorts  of  figures 
having  beep  now  dehvered,  we  may  next  proceed  to  apply 
the©  to  the  sereral  practical  uses  in  life,  as  i'  Hows. 


1Pi3  :  Ki3  : :  2\«c?3  :  2V«<^^  X  C-^-ry  )^==  the  cone  q.mi  ;  therefore 

the  cone  abi  —  the  cone  <5_mi  =•  -^^ad^  — .-Lac^s  x,-— — -)3   ea 

\ad^  h—^adh^  "^-^ah^  is  =  the  conic  frustum  abmq.. 

And  \aa^k  's  =  the  cylinder  ablo. 

Then  the  difierence  of  these  two   is   ladh'^  —  icAs  =  Xak^  X 
(3^— 2A)i  fot  the  spheric  segment  pfn  ;  which  is  the  first  rule. 

Again,  because  pk«  =  fk  X  kh  (cor.  to  theor.  87,  Geom.)  or  rs 

a=!  A  ((i  —  A),  therefore  d  e=s j-  A,  and  Zd  —  2A  =  —  +  A  = 

h  h 

■  I       ;   which  being  substituted  in  the  former  rule,  it  becomes 
h 

^A*  X     ■  =  ^a^^  X  (3r2  -f-  7*2),  which  is  the  2d  rule. 

h 

Note.  By  subtracting  a  segment  from  a  half  sphere,  or  from  another 
«esin«nt,  the  content  of  any  frustum  or  zone  may  be  found. 


LAND 


£  429  j 

LAND  SURVEYING. 

SECTION  I. 

DESCHIPnON  AND  USE  OF  THE  INSTRDMENTS. 

1.  OF  THE  CHAIN. 

LAVD  is  measured  with  a  chain,  called  Gunter's  Chain, 
fl-om  its  inventor,  the  length  of  which  is  4  poles,  or  22  yarcjs, 
or  ^^  feet.  It  consists  of  100  equal  links  ;  and  the  length 
of  each  link  is  therefore  y^/^  of  a  yard,  or  -jVo  of  a  foot,  or 
7*92  inches 

Land  is  estimated  in  acres,  roods,  and  perches.  An  acre 
is  equal  to  10  square  chains,  or  as  much  as  10  chains  in  h  ngth 
and  1  chain  in  breadth.  Or,  in  yards,  it  is  220  X  22  =  4840 
square  yards.  Or,  in  poles,  it  is  40  X  4  =  160  square  poles. 
Or,  in  links,  it  is  1000  X  100  =  100000  square  links  :  these 
being  all  the  same  quantity 

Also,  an  acre  is  divided  into  4  parts  called  roods,  and  a 
rood  into  40  parts  called  perches,  which  are  square  poles  or 
the  square  of  a  pole  of  b\  yards  long,  or  the  square  of  |  of  a 
chain,  or  of  25  links,  which  is  626  square  links.  So  that  the 
divisions  of  land  measure,  will  be  thus  : 

625  sq.  links  =  1  pole  or  perch 
40  perches  =t  1  rood 
4  roods      =  1  acre. 
The  length  of  lines,  measured  with  a  chain,  are  best  set 
down  in  links  as  integers,  every  chain  in  length  being  100 
links  ;  and  not  in  chains  and  decimals.     Therefore   after  the 
content  is  found,  it  will  be  in  square  hnks  ;  then  cut  oif  five 
of  the  figures  on  the   right-hand  for  decimals,  and  the  rest 
will  be  acres.     These  decimals  are  then  multiplied  by  4  for 
roods,  and  the  decimals,  of  these  again  by  40  for  perches. 

Exam.  Suppose  the  length  of  a  rectangular  piece  of  ground 
be  792  links,  and  its  breadth  386  ;  to  find  the  area  in  acres, 
roods,  and  perches. 

792  3-04920 

sas  4 


3960  -19680 

6336  40 

2376  . 

7-87200 


3-04920 


Ahs.  3  acres,  0  roods,  7  perches.  3.  OF 


430  LAND 


2.  OF  THE  PLAIN  TABLE. 

This  instrument  consists  of  a  plain  rectangular  board,  of 
any  convenient  size  :  the  centre /of  which,  when  used,  is  fixed 
by  means  of  screws  to  a  three-legged  stand,  having  a  ball 
and  socket,  or  other  joint,  at  the  top,  by  means  of  which, 
when  the  legs  are  fixed  on  the  ground,  the  table  is  inclined  in 
any  direction. 

To  the  table  belong  various  parts,  as  follow. 

1.  A  frame  of  wood,  made  to  fit  round  its  edges,  and  to 
be  taken  oflf,  for  the  convenience  of  putting  a  sheet  of  paper 
on  the  table.  One  side  of  this  frame  is  usually  divided  into 
equal  parts,  for  drawing  lines  across  the  table,  parallel  or 
perpendicular  to  the  sides  ;  and  the  other  side  of  the  frame 
is  divided  in  360  degrees,  to  a  centre  in  the  middle  of  the 
table  ;  by  means  of  which  the  table  may  be  used  as  a  theo- 
dohte,  &c. 

2  A  magnetic  needle  and  compass,  either  screwed  into  the 
side  of  the  table,  or  fixed  beneath  its  centre,  to  point  out  the 
directions,  and  to  be  a  check  on  the  sights. 

3.  An  index,  which  is  a  brass  two-foot  scale,  with  either  a 
small  telescope,  or  open  sights  set  perpendicularly  on  the 
ends.  These  sights  and  one  edge  of  the  index  are  in  the  same 
plane,  and  that  is  called  the  fiducial  edge  of  the  index. 

To  use  this  instrument,  take  a  sheet  of  paper  which  will 
cover  it,  and  wet  it  to  make  it  expand  ;  then  spread  it  flat  on 
the  table,  pressing  down  the  frame  on  the  edges,  to  stretch 
it  and  keep  it  fixed  there  ;  and  when  the  paper  is  become 
dry,  it  will,  by  contracting  again,  stretch  itself  smooth  and 
flat  from  any  cramps  and  unevenness.  On  this  paper  is  to  be 
drawn  the  plan  or  form  of  the  thing  measured. 

Thus,  begin  at  any  proper  part  of  the  ground,  and  make  a 
point  on  a  convenient  part  fif  the  paper  or  table,  to  repre- 
sent that  place  on  the  ground  ;  then  fix  in  that  point  one  leg 
of  the  compasses,  or  a  fine  steel  pin,  and  apply  to  it  the 
fiducial  edge  of  the  index,  moving  it  round  till  through  the 
sights  you  perceive  some  remarkable  object,  as  the  corner  of 
a  field,  &c.  ;  and  from  the  station-point  draw  a  line  with  the 
point  of  the  compasses  along  the  fiducial  edge  of  the  index, 
which  is  called  setting  or  taking  the  object :  then  set  another 
object  or  corner,  and  draw  its  line  ;  do  the  same  by  another  ; 
and  so  on,  till  as  many  objects  are  taken  as  may  be  thought 
tit.  Then  measure  from  the  station  towards  as  many  of 
the  objects  as  piay  be  necessary,  but  not  more,  taking  the  re- 
<Juisite  offsets  to  corners  or  crooks  in  the  hedges,  laying 
the  measures  down  on  their  respective  lines  on  the  table. 

Then 


SURVEYING.  431 

Then  at  any  convenient  place  measured  to,  fix  the  table  in  the 
same  position,  and  set  the  objects  which  appear  from  that 
place  ;  and  so  on,  as  before.  And  thus  continue  till  the  work 
is  finished,  measuring  such  hnes  only  as  are  necessary,  and 
determining  as  many  as  may  be  by  intersecting  lines  of  direc- 
tion drawn  from  different  stations. 

Of  shifting  the  Paper  on  the  Plain  Table. 

When  one  paper  is  full,  and  there  is  occasion  for  more  ; 
draw  a  line  in  any  manner  through  the  farthest  point  of  the 
last  station  line,  to  which  the  work  can  be  conveniently  laid 
down  ;  then  take  the  sheet  off  the  table,  and  fix  another  on, 
drawing  a  line  over  it,  in  a  part  the  most  convenient  for  the 
rest  of  the  work  ;  then  fold  or  cut  the  old  sheet  by  the  line 
drawn  on  it,  applying  the  edge  to  the  line  on  the  new  sheet, 
and,  as  they  lie  in  that  position,  continue  the  last  station  line 
on  the  new  paper,  placing  on  it  the  rest  of  the  measure,  be- 
ginning at  where  the  old  sheet  left  off.  And  so  on  from  sheet 
to  sheet. 

When  the  work  is  done,  and  you  would  fasten  all  the  sheets 
together  into  one  piece,  or  rough  plan,  the  aforesaid  lines  are 
to  be  accurately  joined  together,  in  the  same  manner  as  when 
the  lines  were  transferred  from  the  old  sheets  to  the  new  ones. 
But  it  is  to  be  noted,  that  if  the  said  joining  lines,  on  the  old 
and  new  sheets,  have  not  the  same  inclination  to  the  side  of 
the  table,  the  needle  will  not  point  to  the  original  degree  when 
the  table  is  rectified  ;  and  if  the  needle  be  required  to  respect 
still  the  same  degree  of  the  compass,  the  easiest  way  of  draVvr- 
ing  the  lines  in  the  same  position,  is  to  draw  them  both  paral- 
lel to  the  same  sides  of  the  table,  by  means  of  the  equal  divi- 
sions marked  on  the  other  two  sides. 

3.  OF  THE  THEODOLITK 

The  theodilite  is  a  brazen  circular  ring,  divided  into  360 
degrees,  &ic.  and  having  an  index  with  sights,  or  a  telescope, 
placed  on  the  centre,  about  which  the  index  is  moveable  ;  also 
a  compass  fixed  to  the  centre,  to  point  out  courses  and  check 
the  sights  ;  the  whole  being  fixed  by  the  centre  on  a  stand  of 
a  convenient  height  for  use. 

In  using  this  instrument,  an  exact  account,  or  field-book,  of 
of  all  measures  and  things  necessary  to  be  remarked  in  the 
plan,  must  be  kept,  from  which  to  make  out  the  plan  on  re- 
turning home  from  the  ground. 

Begin  at  such  part  of  the  ground,  and  measure  in  such  di- 
rections as  are  judged  most  convenient  ;  taking  angles  or  di- 
rections to  objects,  and  measuring  such  distances  as  appear 

necessary, 


43*  LANB 

necessary,  under  the  same  restrictions  as  in  the  use  of  the 
plain  table.  And  it  is  safest  to  fix  the  theodolite  in  the  origi- 
nal pos^ition  at  every  station,  by  means  of  fore  and  back  ob- 
jects, and  the  compass,  exactly  as  in  using  the  plain  table  ; 
registering  the  number  of  degrees  cut  off  by  the  index  when 
directed  to  each  object  ;  and,  at  any  station,  placing  the  index 
at  the  same  degree  as  when  the  direction  towards  that  station 
was  taken  from  the  last  preceding  one,  to  fix  the  theodolite 
there  in  the  original  position. 

The  best  method  of  lying  down  the  aforesaid  lines  of  di- 
rection, is  to  describe  a  pretty  large  circle  ;  then  quarter  it, 
and  lay  on  it  the  several  numbers  of  degrees  cut  off  by  the  in- 
dex in  each  direction,  and  drawing  lines  from  the  centre  to  all 
these  marked  points  in  the  circle.  Then,  by  means  of  a  pa- 
rallel ruler,  draw  from  station  to  station,  hnes  parallel  to  the 
aforesaid  lines  drawn  from  the  centre  to  the  respective  points 
in  the  circumference. 

4.  OF  THE  CROSS. 

The  cross  consists  of  two  pair  of  sights  set  at  right  angles 
to  each  other,  on  a  staff  having  a  sharp  point  at  the  bottom, 
to  fix  in  the  ground. 

The  cross  is  very  useful  to  measure  small  and  crooked  pieces 
of  ground.  The  method  is,  to  measure  a  base  or  chief  line, 
usually  in  the  longest  direction  of  the  piece,  from  corner  to 
corner,  and  while  measuring  it,  finding  the  places  where 
perpendiculars  would  fall  on  this  line,  from  the  several  cor- 
ners and  bends  in  the  boundary  of  the  piece,  with  the  cross, 
by  fixing  it,  by  trials,  on  such  parts  of  the  line,  as  that 
through  one  pair  of  the  sights  both  ends  of  the  line  may 
appear,  and  through  the  other  pair  the  corresponding  bends 
or  corners  ;  and  then  measuring  the  lengths  of  the  said  per- 
pendiculars. 

REMARKS. 

Besides  the  fore-mentioned  instruments,  which  are  most 
commonly  used,  there  are  some  others  :  as, 

The  perambulator,  used  for  measuring  roads,  and  other 
great  distances,  level  ground,  and  by  the  sides  of  rivers.  It 
has  a  wheel  of  V,\  feet,  or  half  a  pole  in  circumference,  by  the 
turning  of  whirJj  the  machine  goes  forward  :  and  the  distance 
measured  is  pointed  out  by  an  index,  which  is  moved  round  by 
clock  work. 

Levels,  with  telescopic  or  other  sights,  are  used  to  find  the 
level  between  place  and  place,  or  how  much  one  place  is 
higher  or  lower  than  another.  And  in  measuring  any  slop- 
ing or  oblique  line,  either  ascending  or  descending,  a  small 

pocket 


SURVEYING.  433 

pocket  level  is  useful  for  showing  how  many  links  for  each 
chain  are  to  be  deducted,  to  reduce  the  line  to  the  horizon- 
tal length. 

An  offset-staff  is  a  very  useful  instrument,  for  measuring 
the  offsets  and  other  short  distances.  It  is  10  links  in  length, 
being  divided  and  marked  at  each  of  the  10  links. 

Ten  small  arrows,  or  rods  of  iron,  or  wood,  are  used  to 
mark  the  end  of  every  chain  length,  in  measuring  lines. 
And  sometimes  pickets,  or  staves  with  flags,  are  set  up  as 
marks  or  objects  of  direction. 

Various  scales  are  also  used  in  protracting  and  measuring 
on  the  plan  or  paper  ;  such  as  plane  scales,  line  or  chords, 
protractor,  compasses,  reducing  scale,  parallel  and  perpen- 
dicular rules,  &c.  Of  plane  scales  there  should  be  several 
sizes,  as  a  chain  in  1  inch,  a  chain  in  f  of  an  inch,  a  chain 
in  ^  an  inch,  &c.  And  of  these,  the  best  for  use  are  those 
that  are  laid  on  the  very  edges  of  the  ivory  scale,  to  mark  off 
distances,  without  compasses. 

SECTION  11. 

THE  PRACTICE  OF  SURVEYING. 

This  part  contains  the  several  works  proper  to  be  done 
in  the  field,  or  the  ways  of  measuring  by  all  the  instruments, 
and  in  all  situations. 

PROBLEM  I. 

To  Measure  a  Line  or  Distance, 

To  measure  a  line  on  the  ground  with  the  chain  :  Having 
provided  a  chain,  with  10  small  arrows,  or  rods,  to  fix  one 
into  the  ground,  as  a  mark,  at  the  end  of  ev^^ry  chain  ;  two 
persons  take  hold  of  the  chain,  one  at  each  end  of  it  ;  and 
all  the  iO  arrows  are  taken  by  one  of  them,  who  goes  fore- 
most, and  is  called  the  leader  ;  the  other  being  called  the 
follower,  for  distinction's  sake. 

A  picket,  or  station-staff  being  set  up  in  the  direction  of 
the  line  to  be  measured,  if  there  do  not  appear  some  marks 
naturally  in  that  direction,  they  measure  straight  towards  it, 
the  leader  fixing  down  an  arrow  at  the 'end  of  every  chain, 
which  the  follower  always  takes  up,  as  he  comes  at  it,  till  all 
the  ten  arrows  are  used.  They  are  then  all  returned  to  the 
leader,  to  use  over  again.  And  thus  the  arrows,  are  chang- 
ed from  the  one  to  the  other  at  every  10  chains'  length, 
till  the  whole  line  is  finished  ;  then  the  number  of  changes 

Tct.  I.  h%  of 


434  LAND 

of  the  arrows  shows  the  number  of  tens,  to  which  the  fol- 
lower adds  the  arrows  he  holds  in  his  hand,  and  the  number 
of  hnks  of  another  chain  over  to  the  mark  or  end  of  the 
line.  So,  if  there  have  been  3  changes  of  the  arrows,  and 
the  follower  hold  6  arrows,  and  the  end  of  the  line  cut  off 
45  links  more,  the  whole  length  of  the  line  is  set  down  in  links 
thus,  3646 

When  the  ground  is  not  level,  but  either  ascending  or  de- 
scending ;  at  every  chain  length,  lay  the  offset-staff,  or  link- 
staff,  down  in  the  slope  of  the  chain,  on  which  lay  the  small 
pocket  level,  to  show  how  many  links  or  parts  the  slope  line 
is  longer  than  the  true  level  one  ;  then  draw  the  chain  for- 
ward so  many  links  or  parts,  which  reduces  the  line  to  the 
horizontal  direction. 

PROB!uEM  II. 

To  take  Angles  and  Bearings, 

Let  b  and  c  be  two  objects,  or 
two  pickets  set  up  perpendicular; 
and  let  it  be  required  to  take  their 
bearings,  or  the  angles  formed  be- 
tween them  at  any  station  a. 


!.   With  the  Plain  Table, 

The  table  being  covered  with  a  paper,  and  fixed  on  its 
stand ;  plant  it  at  the  station  a,  and  fix  a  fine  pin,  or  a  foot 
of  the  compasses,  in  a  proper  point  of  the  paper,  to  repre-  ' 
sent  the  place  a  :  Close  by  the  side  of  this  pin  lay  the  fiducial 
edge  of  the  index,  and  turn  it  about,  still  touching  the  pin, 
till  one  object  b  can  be  seen  through  the  sights  :  then  by  the 
fiducial  edge  of  the  index  draw  a  line.  In  the  same  manner 
draw  another  line  in  the  direction  of  the  other  object  c. 
And  it  is  done. 

2.  With  the  Theodolite,  ^c. 

Direct  the  fixed  sights  along  one  of  the  lines,  as  ab,  by 
turning  the  instrument  about  till  the  mark  b  is  seen  through 
these  sights  ;  and  there  screw  the  instrument  fast.  Then 
turn  the  moveable  index  round,  till  through  its  sights  the 
other  mark  c  is  seen.  Then  the  degrees  cut  by  the  index, 
on  the  graduated  limb  or  ring  of  the  instrument,  show  the 
quantity  of  the  angle. 

3.  With 


SURVEYING,  43$ 

3.   With  the  Magnetic  Needle  and  Compass. 

Turn  the  instrament  or  compass  so,  that  the  north  end 
©f  the  needle  point  to  the  flower-de-luce.  Then  direct  the 
sights  to  one  mark  as  b,  and  note  the  degrees  cut  by  the 
needle.  Next  direct  the  sights  to  the  other  mark  c,  and 
note  again  the  degrees  cut  by  the  needle.  Then  their  Sum 
or  difference,  as  the  case  may  be,  will  give  the  quantity  of 
the  angle  bag. 

4.  By  Measurement  with  the  Chaiuy  «$*c. 

Measure  one  chain  length,  or  any  other  length,  along 
both  directions,  as  to  b  and  c.  Then  measure  the  distance 
b^  c.  and  it  is  done. — This  is  easily  transferred  to  paper,  by 
making  a  triangle  Abe  with  these  three  lengths,  and  then 
measuring  the  angle  a. 

PROBLEM  ni. 

To  Survey  a  Triangular  Field  abc. 
'  I.  By  the  Chain. 


AP  794 
AB  1321 
PC  826 


c 


A  P     B 

Having  set  up  marks  at  the  corners,  which  is  to  be  done 
in  all  cases  where  tuere  are  not  marks  naturally  ;  measure 
with  the  chain  from  a  to  p,  where  a  perpendicular  would 
fall  from  the  angle  c,  and  set  up  a  mark  at  p,  noting  down 
the  distance  ap.  Then  complete  the  distance  ab,  by  mea- 
suring from  p  to  B.  Having  set  down  this  measure,  return 
to  p,  aud  measure  the  perpendicular  pc.  And  thus,  having 
the  base  and  perpendicular,  the  area  from  them  is  easily 
found.  Or  having  the  place  p  of  the  perpendicular,  the 
triangle  is  easily  constructed. 

Or,  measure  all  the  three  sides  with  the  chain,  and  note 
them  down.  From  which  the  content  is  easily  found,  or  the 
figure  is  constructed. 

2.  By  taking  som£  of  the  Angles. 

Measure  two  sides  ab,  ac,  and  the  angle  a  between  them. 
Or  measure  one  side  ab,  and  the  two  adjacent  angles  a  and 
B.  From  either  of  these  ways  the  figure  is  easily  planned  ; 
then  by  measuring  the  perpendicular  cp  on  the  plan,  and 
multiplying  it  by  half  ab,  the  content  is  found. 

PBGBLBM 


436 


LAND 

PROBLEM  IV. 
To  Measure  a  Four-sided  Fields 


AE214 

AF  362 
AC  692 


210  DE 

306  BF 


1.  By  the  Chain, 


Measure  along  one  of  the  diaponaU,  as  ac  ;  and  either 
the  two  perpendiculars  de,  bf,  as  in  the  last  problem  ;  or 
else  the  sides  ab,  bc,  cd,  da.  From  either  of  which  the 
figure  ma}?  be  planned  and  computed  as  before  directed. 

Otherwise  by  the  Chain. 


AP 

no 

352  PC 

Aq, 

745 

595  QD 

AB 

1110 

.AP 

Measure,  on  the  longest  side,  the  distances  ap, 
and  the  perpendiculars  pc,  qd. 

2.  15y  taking  some  of  the  Angles. 

Measure  the  diagonal  ac  (see  the  last  fig.  but  one),  and 
the  angles  cab,  cad,  acb,  acd. — Or  measure  the  four  sides, 
and  any  one  of  the  angles,  as  bad. 

Or  thus. 
ab     48G 
BC     394 
CD     410 
DA     462 
BAD  78<>  35' 
PROBLEM  V. 
To  Survey  any  Fiela  by  the  Chain  only. 
Having  set  up  marks  at  the  corners,  where  necessary,  of 
the  proposed  field  abcdefg,  walk  over  the  ground,  and  con- 
sider how  it  can  best  be  divided  in  triangles  and  trapeziums  ; 
and  measure  them  separately,  as  in  the  last  two  problems. 
Thus,  the  following  figure  is  divided  into  the  two  trapeziums 
ABCG,  GDEF,  and  the  triangle  gcd.     Then,  in  the  first  tra-- 
pezium,   beginning  at  a,  measure  the  diagonal  ac,   and  the 

two 


Thus. 

AC  591 

CAB   37° 

20' 

CAD   41 

15 

ACB   72 

25 

ACD   54 

40 

SURVEYING. 


437 


two  perpendiculars  cm,  sn.  Then  the  base  cc,  and  the 
perpendicular  Dq.  Lastly,  the  diagonal  df,  and  the  two 
perpendiculfirs  pE,  og.  All  which  measures  write  agjainst 
the  corresponding  parts  of  a  rough  figure  drawn  to  resemble 
the  figure  surveyed,  or  set  them  down  in  any  other  form  you 
choose. 


Thus. 

Am 

135  130 

mo 

An 

410 

180 

nB 

AC 

650 

cq 

162 

230 

qo 

CG 

440 

FO 

237 

120 

OG 

Fp 

288 

80 

PE 

FD 

520 

Or  thus. 


Measure  all  the  sides  ab,  bc,  cd,  de,  ef,  fg,  ga  ;  and  the 
diagonals  ac,  cg,  gd,  df.  ^ 

Otherwise, 

Many  pieces  of  land  may  be  very  well  surveyed,  by  mea- 
suring any  base  line,  either  within  or  without  them,  with  the 
perpendiculars  let  fall  on  it  from  every  comer.  For  they  arc 
'  by  those  means  divided  into  several  triangles  and  trapezoids, 
all  whose  parallel  sides  are  perpendicular  to  the  base  hue  ; 
and  the  sum  of  these  triangles  and  trapeziums  will  be  equal 
to  the  figure  proposed  if  the  base  line  fall  within  it  ;  if  not 
the  sum  of  the  parts  which  are  without  being  taken  from  the 
sum  of  the  whole  which  are  both  within  and  without,  will  leave 
the  area  of  the  figure  proposed. 

In  pieces  that  are  not  very  large,  it  will  be  sufficiently  ex- 
act to  find  the  points,  in  the  base  line,  where  the  several  per- 
pendiculars will  fall,  by  means  of  the  cross^  or  even  by  judg- 
ing by  the  eye  only,  and  from  thence  measuring  to  the  corners 
for  the  lengths  of  the  perpendiculars. — And  it  will  be  most 
convrenient  to  draw  the  line  so  as  that  all  the  perpendiculars 
may  fall  within  the  figure. 

Thus  in  the  following  figure,  beginning  at  a,  and  measuring 
along  the  line  ag,  the  distances  and  perpendiculars  on  the 
right  and  left  are  as  below. 

Ab 


438 


LANB 


Ab 

316 

350  bB 

AC 

440 

70  cc 

Ad 

686 

320  do 

AC 

610 

60  eE 

a£ 

990 

470  fF 

Ag 

1020 

0 

PROBLEM  VI. 

To  Measure  the  Offsets. 
Ahiklmn  being  a  crooked  hedge,  or  brook,  &c.  From  A 
measure  in  a  straight  direction  along  the  side  of  it  to  b.  .'^nd 
in  measuring  along  this  line  ab,  observe  when  you  are  direct- 
ly opposite  any  bends  or  corners  of  the  boundary,  as  at  c,  d, 
e,  &c.  ;  and  from  these  measure  the  perpendicular  offsets  ch, 
di,  &c.  with  the  oflfset-staff,  if  they  are  not  very  large,  other- 
wise with  the  cham  itself ;  and  the  work  is  done.  The  regis- 
ter, or  field  book,  may  be  as  follows  : 


Uffs. 

left. 

Base  line  ab 

0 

0        A 

ch 

62 

45     AC 

di 

84 

220     Ad 

ek 

70 

340     AC 

fl 

98 

610  ■  Af 

:^m 

67 

634     Ag 

Bn 

91 

786     AB 

AC 


PROBLEM  Vir. 

To  survey  any  Field  with  the  Plain  Table, 
1.  From  one  Station. 


Plant  the  table  at  any  angle  as 
G,  from  which  all  the  other  angles, 
or  marks  set  up,  can  be  seen  ;  turn 
the  table  about  till  the  needle  point 
to  the  flower-de-luce  :  and  there 
screw  it  fast.  Make  a  point  for 
c  on  the  paper  on  the  table,  and 
lay  the  edge  of  the  index  to  c, 
turning  it  about  c  till  through  the  A  B 

sights  you  see  the  mark  d  :  and  by  the  edge  of  the  index 
draw  a  dry  or  obscure  line  :  then  measure  the  distance  cd, 
and  lay  that  distance  down  on  the  line  cd.  Then  turn  the 
index  about  the  point  c,  till  the  mark  e  be  seen  through  the 

'  'its, 


SURVEYING.  43» 

sights,  by  which  draw  a  line,  and  measure  the  distance  to  e, 
laying  it  on  the  hne  from  c  to  e.  In  like  manner  determine 
the  positions  of  ca  and  cb,  by  turning  the  sights  successively  to 
A  and  B  ;  and  lay  the  lengths  of  those  lines  down.  Then 
connect  the  points,  by  drawing  the  black  lines  cd,  de,  ea,  ab^ 
Bc,  for  the  boundaries  of  the  field. 

2.  From  a  Station  Within  the  Field. 

When   all  the   other  parts  cannot  - 

be  seen  from  one  angle,  choose  some 
place  0  within,  or  even  without,  if 
more  convenient,  from  which  the 
other  parts  can  be  seen.  Plant  the 
table  at  0,  then  fix  it  with  the  needle 
north,  and  mark  the  point  0  on  it. 
Apply    the  index  successively  to  0,  _ 

turning  it  round   with   the  sights  to  A 

each  angle,  a,  b,  c,  d,  e,  drawing  dry  lines  to  them  by  the  edge 
of  the  index  ;  then  measuring  the  distance  oa,  ob,  &,c.  and 
laying  them  down  on  those  lines.  Lastly,  draw  the  bounda- 
ries AB,  bc,  CD,  DE,  EA.. 

.  3.  By  going  Round  the  Figure. 

When  the  figure  is  a  wood,  or  water,  or  when  from  some 
other  obstruction  you  cannot  measure  lines  across  it  :  begia 
at  any  point  a,  and  measure  around  it  either  within  or 
without  the  figure,  and  draw  the  directions  of  all  the  sides, 
thus  ;  Plant  the  table  at  a  ;  turn  it  with  the  needle  to  the 
north  or  flower-de-luce  ;  fix  it,  and  mark  the  point  a.  Apply 
the  index  to  a,  turning  it  till  you  can  see  the  point  e,  and 
there  draw  a  line  :  then  the  point  b,  and  there  draw  a  line  ; 
then  measure  these  lines,  and  lay  them  down  from  a  to  e  and 
B.  Next  move  the  table  to  b,  lay  the  index  along  the  line 
AB,  and  turn  the  table  about  till  you  can  see  the  mark  a,  and 
screw  fast  the  table  ;  in  which  position  also  the  needle  will 
again  point  to  the  flower-de-luce,  as  it  will  do  indeed  at  every 
station  when  the  table  is  in  the  right  position.  Here  tura 
the  index  about  b  till  through  the  sights  you  see  the  mark  c  ; 
there  draw  a  line,  measure  bc,  and  lay  the  distance  on  that 
line  after  you  have  set  down  the  table  at  c.  Turn  it  then 
again  into  its  proper  position,  and  in  like  manner  find  the  next 
line  CD.  And  so  on  quite  around  by  e,  to  a  again.  Then  the 
proof  of  the  work  will  be  the  joining  at  a  ;  fdr  if  the  work  be 
all  right,  the  last  direction  ea  on  the  ground,  will  pass  exactly 
through  the  point  a  on  the  paper  ;  and  the  measured  distance 
will  also  reach  exactly  to  a.  If  these  do  not  coincide,  or  nearly 
so,  some  error  has  b^ea  cemBOiitted,  aqd  the  work  must  be  ex- 
aiQined  over  again. 

PROBLEM 


440  -  LAND 

PROBLEM  VIII. 

To  Survey  a  Field  with  the  Theodolite,  ^c, 
1 .  From  One  Point  or  Station. 
When  all  the  angles  can  be  seen  from  one  point,  as  the 
angle  c  (first  fig.  to  last  prob.)  place  the  instrnmenf  at  c.  and 
turn  it  about,  till  through  the  fixed  sights  you  see  the  mark 
B,  and  there  fix  it.  Then  turn  the  moveable  index  about 
till  the  mark  a  be  seen  through  the  sights,  and  note  the  de- 
grees cut  on  the  instrument.  Mextturn  the  index  successively 
to  E  and  D,  noting  the  degrees  cut  off  at  each  ;  \^hich  gives  ill 
the  angles  bca.  bce,  bcd.  Lastly  measure  the  lines  cb,  ca,  ce, 
CD  ;  and  enter  the  measures  in  a  field-book,  or  rather  against 
the  corresponding  parts  of  a  rough  figure  drawn  by  guess  to 
resemble  the  field. 

1 .  From  a  point  Within  or  Without, 

Plant  the  instrument  at  0  (last  fig.)  and  turn  it  about  till  the 
fixed  sights  point  to  any  object,  as  a  ;  and  there  screw  it  fast. 
Then  turn  the  moveable  index  round  till  the  sights  point  suc- 
cessively to  the  other  points  e,  d,  c,  b,  noting  the  degrees  cut 
off  at  each  of  them  ;  which  gives  all  the  angles  round  the  point 
0.  Lastly  measure  the  distances  oa,  ob,  oc,  od,  oe,  noting 
them  down  as  before,  and  the  work  is  done. 

3.  By  going  Round  the  Field. 

By  measuring  round,  either 
within  or  without  the  field,  pro- 
ceed thus.  Having  set  up  marks 
at  B,  c,  kc.  near  the  corners  as 
usual,  plant  the  instrument  at 
any  point  a,  and  turn  it  till  the 
fixed  index  be  in  the  direction 
AB,  and  there  screw  it  fast :  then 
turn  the  moveable  index  to  the 
direction  af  ;  and  the  degrees  cut  off  will  be  the  angle  a. 
Measure  the  line  ab,  and  plant  the  instrument  at  b,  and 
there  in  the  same  manner  observe  the  angle  a.  Then  measure 
Bc,  and  observe  the  angle  c.  Then  measure  the  distance  cd, 
and  take  the  angle  d.  'Phen  measure  de,  and  take  t*^e  angle  e. 
Then  measure  ef,  and  take  the  angle  f.  And  lastly  measure 
the  distance  fa. 

To  prove  the  work  ;  add  all  the  inward  angles  a,  b,  c, 
&c.  together  ;  for  ,when  the  work  is  right,  their  sum  will  be 
equal  to  twice  as  many  right  angles  as  the  figure  has  sides, 
wanting  4  right  angks.  But  when  there  is  an  angle,  as  f, 
that  bends  inwards,  and  you  measure  the  external   angle, 

which 


SURVEYING. 


441 


which  is  less  than  two  right  angles,  subtract  it  from  4  right 
angles,  or  360  degrees,  to  give  the  internal  angle  greater 
than  a  semicircle  or  180  degrees. 

Olherwise, 
Instead  of  observing  the  internal  angles,  we  may  take  the 
external  angles,  formed  without  the  figure  by  producing  the 
sides  farther  out.  And  in  this  case,  when  the  work  is  right, 
their  sum  altogether  will  be  equal  to  360  degrees.  But  when 
one  of  them,  as  f,  runs  inwards,  subtract  it  from  the  sum  of 
the  rest,  to  leave  360  degrees. 

PROBLEM  IX. 

To  Survey  a  Field  with  Crooked  Hedges,  S^c. 

WITH  any  of  the  instruments,  measure  the  lengths  and 
positions  of  imaginary  lines  running  as  near  the  sides  of  the 
field  as  you  can  ;  and,  in  going  along  them,  measure  the 
offsets  in  the  manner  before  taught  ;  then  you  will  have  the 
plan  on  the  paper  in  using  the  plain  table,  drawmg  the 
crooked  hedges  through  the  ends  of  the  offsets  ;  but  in  sur- 
veying with  the  theodolite,  or  other  instrument,  set  dowa 
the  measures  properly  in  a  field-book,  or  memorandum- 
book,  and  plan  them  after  returning  from  the  field,  by  laying 
down  all  the  lines  and  angles. 


So,  in  surveying  the  piece  abcde,  set  up  marks  a,  b,  c,  d, 
dividing  it  so  as  to  have  as  few  sides  as  may  be.  Then  begin  at 
any  station,  a,  and  measure  the  lines  ab,  be,  cd,  da,  taking 
their  positions,  or  the  angles  a,  b,  c,  d  ;  and,  in  going  along 
the  lines,  measure  all  the  offsets,  as  at  m,  n,  o,  p,  &jc.  along 
every  station-line. 

And  this  is  done  either  within  the  field,  or  without,  as 
may  be  most  convenient.  When  there  are  obstructions 
within,  as  wood,  water,  hills,  &c.  then  measure  without,  as 
in  the  next  following  figure. 

^OL.  I.  57  PROBLEM 


442 


LAND 


PROBLEM  X. 

To  Survey  a  Field,  or  any  other  Things  by  Two  Stations. 

This  is  performed  by  choosing  two  stations  from  which 
all  the  marks  and  objects  can  be  seen  ;  then  measuring  the 
distance  between  the  stations,  and  at  each  station  taking  the 
angles  formed  by  every  object  from  the  station  line  or  dis- 
tance. 

The  two  stations  may  be  taken  either  within  the  bounds, 
or  in  one  of  the  sides,  or  in  the  direction  of  two  of  the  objects, 
or  quite  at  a  distance  and  without  the  bounds  of  the  objects 
or  part  to  be  surveyed. 

In  this  manner,  not  only  grounds  may  be  surveyed,  with- 
out even  entering  them,  but  a  map  may  be  taken  of  the 
principal  parts  of  a  county,  or  the  chief  places  of  a  town, 
or  any  part  of  a  river  or  coast  surveyed,  or  any  other  inacces- 
sible objects  ;  by  taking  two  stations,  on  two  towers,  or  two 
hills,  or  such-like. 


PROBLEM  XL 

To  Survey  a  Large  Estate. 
If  the  estate  be  very  large,  and  contain  a  great  number  of 
fields,  it  cannot  well    be    done  by  surveying  all   the  fields 

singly 


SURVEYING.  443 

singly,  and  then  putting  them  together  ;  nor  can  it  be  done 
by  taking  all  the  angles  and  boundaries  that  enclose  it.  For 
in  these  cases,  any  small  errors  will  be  so  much  increased,  as 
to  render  it  very'raach  distorted.     But  proceed  as  below. 

1.  Walk  over  the  estate  two  or  three  times,  in  order  to 
get  a  perfect  idea  of  it,  or  till  you  can  keep  the  figure  of  it 
pretty  well  in  mind.  And  to  help  your  memory,  draw  an 
eye-draught  of  it  on  paper,  or  at  least  of  the  principal  parts 
of  it,  to  guide  you  ;  -setting  the  names  within  the  fields  in 
that  draught. 

2.  Choose  two  or  more  eminent  places  in  the  estate,  for 
stations,  from  which  all  the  principal  parts  of  it  can  be  seen  : 
selecting  these  stations  as  far  distant  from  one  another  as 
convenient. 

3.  Take  such  angles,  between  the  stations,  as  you  think 
necessary,  and  measure  the  distances  from  station  to  station, 
always  in  a  right  line  :  .these  things  must  be  done,  till  you 
get  as  m^ny  angles  and  lines  as  are  sufficient  for  determiumg 
all  the  points  of  station.  And  in  measuring  any  of  these 
station-distances,  mark  accurately  where  these  lines  meet 
with  any  hedges,  ditches,  roads,  lanes,  paths,  rivulets,  &c.  ; 
and  where  any  remarkable  object  is  placed,  by  measuring  its 
distance  from  the  station-line  ;  and  where  a  perpendicular 
from  it  cuts  that  line.  And  thus  as  you  go  along  any  main 
station-line,  takft  offsets  to  the  ends  of  all  hed<res,  and  to  any 
pond,  house,  mill,  bridge,  &.c.  noting  every  thing  down  that 
is  remarkable. 

4.  As  to  the  inner  parts  of  the  estate,  they  must  be  deter- 
mined, in  hke  manner,  by  new  station-lines  :  for,  aft^^r  the 
main  stations  are  determined,' and  every  thing  adjoining  to 
them,  then  the  estate  must  be  subdivided  into  two  or  three 
parts  by  new  station-lines  ;  taking  inner  stations  at  proper 
places,  where  you  can  have  the  best  view.  Mv^^asure  these 
station-lines  as  you  did  the  first,  and  all  their  intersections 
with  hedges  and  offsets  to  such  objects  as  appear.  Then 
proceed  to  survey  the  adjoining  fields,  by  taking  the  angles 
that  the  sides  make  with  the  station-line,  at  the  intersections, 
and  measuring  the  distances  to  each  corner,  from  the  inter- 
sections. For  the  station- lines  will  be  the  bases  to  all  the 
future  operations  ;  the  situation  of  all  parts  being  entirely 
dependent  on  them  ;  and  therefore  they  should  be  taken  of 
as  great  length  as  possible  ;  and  it  is  best  for  them  to  run 
along  some  of  the  hedges  or  boundaries  of  one  or  more  fields, 
or  to  pass  through  some  of  their  angles.     All  things   being 

.determined  for  these  stations,  you  must  take  more  inner  sta- 
tions, and  continue  to  divide  and  subdivide  till  at  last  you 
r.ome  to  single  fields  ;  repeating  the  same  work  for  the  inner 

stations 


444  LAND 

stations  as  for  the  outer  ones,  till  all  is  done  ;  and  close  the 
work  as  often  as  you  can,  and  in  as  few  lines  as  possible. 

5.  An  estate  may  be  so  situated  that  the  whole  cannot  be 
surveyed  together  ;  because  one  part  of  the  estate  cannot  be 
seen  from  another.  In  this  case,  you  may  divide  it  into 
three  or  four  parts,  and  survey  the  parts  separately,  as  if 
they  were  lands  belonging  to  different  persons  ;  and  at  last 
join  them  together. 

6.  As  it  is  necessary  to  protract  or  lay  down^the  work  as 
you  proceed  in  it,  you  must  have  a  scale  of  a  due  length  to 
do  it  by.  To  get  such  a  scale,  measure  the  whole  length  of 
the  estate  in  chains  ;  then  consider  how  many  inches  long 
the  map  is  to  be  ;  and  from  these  will  be  known  how  many 
chains  you  must  have  in  an  inch  ;  then  make  the  scale  accor- 
dingly, or  cho6se  one  already  made. 

PROBLEM  Xn. 

To  Survey  a  County,  or  large  Trad  of  Land. 

1.  Choose  two,  three,  or  four  eminent  places,  for  stations  : 
such  as  the  tops  of  high  hills  or  mountains,  towers,  or  church 
steeples  which  may  be  seen  from  one  another  ;  from  which 
most  of  the  towns  and  other  places  of  note  may  also  be  seen  ; 
and  so  as  to  be  as  far  distant  from  one  another  as  possible. 
On  these  places  raise  beacons,  or  long  poles,  with  tiags  of 
different  colours  flying  at  them,  so  as  to  be  visible  from  all 
the  other  stations. 

2.  At  all  the  places  which  you  would  set  down  in  the 
map,  plant  long  poles,  with  flags  at  them  of  several  colours, 
to  distinguish  the  places  from  one  another;  fixinav.  them  on 
the  tops  of  church  steeples,  or  the  tops  of  houses  ;  or  in  the 
centres  of  smaller  towns  and  villages. 

These  marks  then  being  set  up  at  a  convenient  number  of 
places,  and  such  as  may  be  seen  from  both  stations  ;  go  to 
one  of  these  stations,  and,  with  an  instrument  to  take  angles, 
,  standing  at  that  station,  take  all  the  angles  between  the  other 
station  and  each  of  these  marks.  Then  go  to  the  other 
station  and  take  all  the  angles  between  the  first  station  and 
each  of  the  former  marks,  setting  them  down  with  the  others, 
each  against  its  fellow  with  the  same  colour.  You  may,  if 
convenient,  also  take  the  angles  at  some  third  station,  which 
may  serve  to  prove  the  work,  if  the  three  lines  intersect  in 
that  point  where  any  mark  stands.  The  marks  must  stand  till 
the  observations  are  finished  at  both  stations  ;  and  then  they 
may  be  taken  down,  and  set  up  at  new  places.  .  The  same 
operations  must  be  performed,  at  both  stations,  for  these 
new  places  ;  and  the  like  for  others.     The  instrument   for 

taking 


SURVEYING.  445 

taking  angles  must  loe  an  exceeding  good  one,  made  on  pur- 
pose with  telescopic  sights,  and  of  a  good  length  of  radius. 

3.  And  though  it  be  not  absolutely  necessary  to  mea^su^e 
any  distance,  because  a  stationary  line  being  laid  down  from 
any  scale,  all  the  other  lines  will  be  proportional  to  it  ;  yet  it 
is  l3etter  to  measure  some  of  the  lines,  to  ascertain  the  dis- 
tances of  places  in  miles,  and  to  know  how  many  geometrical 
miles  there  are  in  any  length  ;  as  also  from  thence  to  make  a 
scale  to  measure  any  distance  in  miles.  In  measuring  any 
distance,  it  will  not  ,be  exact  enough  to  go  along  the  high 
roads  ;  which,  by  reason  of  their  turnings  and  windings,  hard- 
ly ever  he  in  a  right  line  between  the  stations  ;  which  must 
cause  endless  reductions,  and  require  great  trouble  to  make 
it  a  right  line  ;  for  which  reason  it  can  never  be  exact.  But 
a  better  way  is  to  measure  in  a  straight  line  with  a  chain,  be- 
tween station  and  station,  over  hills  and  dales,  or  level  fields, 
and  all  obstacles  Only  in  case  of  water,  woods,  towns,  rocks, 
banks,  &c.  where  we  cannot  pass,  such  parts  of  the  line  must 
be  measured  by  the  methods  of  inaccessible  distances  ;  and 
besides  allowing  for  ascents  and  decents,  when  they  are  met 
with.  A  good  compass,  that  shows  the  bearing  of  the  two 
stations,  will  always  direct  us  to  go  straight,  when  the  two 
stations  cannot  be  seen  ;  and  in  the  progress,  if  we  can  go 
straight,  offsets  may  be  taken  to  any  remarkable  places, 
likewise  noting  the  intersection  of  the  station-line  with  all 
roads,  rivers,  &c. 

4.  From  all  the  stations,  and  in  the  whole  progress,  we 
must  be  very  particular  in  observing  sea  coasts,  river-mouths, 
towns,  castles,  houses,  churches,  mills,  trees,  rocks,  sands, 
roads,  bridges,  fords,  ferries,  woods,  hills,  mountains,  rills, 
brooks,  parks,  beacons,  sluices,  floodgates,  locks,  &c.  and  in 
general  every  thing  that  is  remarkable. 

5.  After  we  have  done  with  the  first  and  main  station-lines, 
which  command  the  whole  county  ;  we  must  then  take  inner 
stations  at  some  places  already  determined  ;  which  will  di- 
vide the  whole  into  several  partitions  :  and  from  these  stations 
we  must  determine  the  places  of  as  many  of  the  remaining 
towns  as  we  can.  And  if  any  remain  in  that  part,  we  must 
take  more  stations,  at  some  places  already  determined,  from 
which  we  may  determine  the  rest.  And  thus  go  through  all 
the  parts  of  the  county,  taking  station  after  station,  till  we 
have  determined  the  whole.  And  in  general  the  station-dis- 
tances must  always  pass  through  such  remarkable  points  as 
have  been  determined  before,  by  the  former  stations. 

PROBLEM 


446 


LAND 


PROBLEM  XUI. 

To  Survey  a  Town  or  City. 

This  may  be  done  with  any  of  the  instruments  for  taking 
angles,  but  best  of  all  with  the  plain  table,  where  every  mi- 
nute part  is  drawn  while  in  sight.  Instead  of  the  common 
surveying  or  Gunter's  chain,  it  will  be  best,  for  this  purpose, 
to  have  a  chain  50  feet  long,  divided  into  50  links  of  one  foot 
each,  and  an  offset-staff  of  10  feet  long. 

Begin  at  the  meeting  of  two  or  more  of  the  principal 
streets,  through  which  we  can  have  the  longest  prospects,  to 
get  the  longest  station-lines  :  there  having  fixed  the  instru- 
ment, draw  lines  of  direction  along  those  streets,  using  two 
men  as  marks,  or  poles  set  in  wooden  pedestals,  or  perhaps 
some  remarkable  places  in  the  houses  at  the  farther  ends, 
as  windows,  doors,  corners,  &c  Measure  these  lines  with 
the  chain,  taking  offsets  with  the  staff,  at  all  corners  of 
streets,  bendings,  or  windings,  and  to  all  remarkable  things,  as 
churches,  markets,  halls,  colleges,  eminent  houses,  &c.  Then 
remove  the  instrument  to  another  station,  along  one  of  these 
lines  ;  and  there  repeat  the  same  process  as  before.  And  so 
on  tillthe  whole  is  finished. 


Thus,  fi3f  the  instrument  at  a,  and  draw  lines  in  the  direc- 
tion of  all  the  streets  meeting  there  ;  then  measure  ab,  noting^ 
the  street  on  the  left  at  m.  At  the  second  station  b,  draw  the 
directions  of  the  streets  meeting  there  ;  and  measure  from  b 
to  c,  noting  the  places  of  the  streets  at  n  and  o  as  you  pass  by 
them.  At  the  third  station  c,  take  the  direction  of  all  the 
streets  meeting  there,  and  measure  cd.  At  d  do  the  same, 
and  measure  de,  noting  the  place  of  the  cross  streets  at  p.  And 
in  this  manner  go  through  all  the  principal  streets  This  done, 
proceed  to  the  smaller  and  intermediate  streets  ;  and  lastly  to 
the  lanes,  alleys,  courts,  yards,  and  every  part  that  it  may  be 
thought  proper  to  represent  in  the  plan. 


SURVEYING.  447 

PRCWBLEM  XIV. 

To  lay  down  the  Plan  of  any  Survey. 

Ip  the  survey  was  taken  with  the  plain  table  we  have  a  rough 
plan  of  it  already  on  the  paper  which  covered  the  table.  But 
if  the  survey  was  with  any  other  instrument,  a  plan  of  it  is  to 
be  drawn  from  the  rneasuies  that  were  taken  in  the  survey  ; 
and  first  of  all  a  rough  planon  paper. 

To  do  this,  you  must  have  a  set  of  proper  instruments,  for 
laying  down  both  lines  and  angles,  kc.  ;  as  scales  of  various 
sizes  (the  more  of  them,  and  the  more  accurate,  the  better), 
scales  of  chords,  protractors,  perpendicular  and  parallel  rulers, 
&c.  Diagonal  scales  are  best  for  the  lines,  because  they  ex- 
tend to  three  figures,  or  chains,  and  links,  which  are  100  parts 
of  chains.  But  in  using  the  dia^ronal  scale,  a  pair  of  compasses 
must  be  employed,  to  take  off  the  lengths  of  the  principal 
lines  very  accurately.  But  a  scale  with  a  thin  edge  divided,  is 
much  readier  for  laying  down  the  perpendicular  offsets  to 
crooked  hedges,  and  for  marking  the  places  of  those  offsets  on 
the  station-line  ;  which  is  done  at  only  one  application  of  the 
edge  of  the  scale  to  that  line,  and  then  pricking  off  all  at  once 
the  distances  along  it.  Angles  are  to  be  laid  down  either  with 
a  good  scale  of  chords,  which  is  perhaps  the  most  accurate 
way,  or  with  a  large  protractor,  which  is  much  readier  when 
many  angles  are  to  be  laid  down  at  one  point,  as  they  are  prick- 
ed off  all  at  once  round  the  edge  of  the  protractor. 

In  general,  all  lines  and  angles  must  be  laid  down  on  the 
plan  in  the  sam^  order  in  which  they  were  measured  in  the 
field,  and  in  which  they  are  written  in  the  field-book  ;  lay- 
ing down  first  the  angles  for  the  position  of  lines,  next  the 
lengths  of  the  lines,  with  the  places  of  the  offsets,  and  then 
the  lengths  of  the  offsets  themselves,  all  with  dry  or  obscure 
lines  ;  then  a  black  line  drawn  through  the  extremities  of  all 
the  offsets,  will  be  the  hedge  or  bounding  line  of  the  field, 
&c.  After  the  principal  bounds  and  lines  are  laid  down,  and 
made  to  fit  or  close  properly,  proceed  next  to  the  smaller 
objects,  till  you  have  entered  every  thing  that  ought  to  appear 
in  the  plan,  as  houses,  brooks,  trees,  hills,  gates,  stiles,  roads, 
lanes,  mills,  bridges,  woodlands,  &c.  &c. 

The  north  side  of  a  map  or  plan  is  commonly  placed 
uppermost,  and  a  meridian  is  some  where  drawn,  with  the 
compass  or  flower-de-luce  pointing  north.  Also,  in  a  vacant 
part,  a  scale  of  equal  parts  or  chains  is  drawn,  with  the  title 
.of  the  map  in  conspicuous  characters,  and  erat)ellished  with 
a  compartment.  Hills  are  shadowed  to  distinguish  them  in 
the  map.  Colour  the  hedges  vv^th  different  colours  ;  repre- 
sent 


448  LAND 

sent  billy  grounds  by  broken  bills  and  valleys  ;  draw  single 
dotted  lines  for  foot-paths,  and  double  ones  for  horse  or  car- 
riage roads.  Write  the  name  of  each  field  and  remarkable 
place  within  it,  and,  if  you  choose,  its  content  in  acres,  roodSj 
and  perches. 

In  a  very  large  estate,  or  a  county,  draw  vertical  and  ho- 
rizontal lines  through  the  map,  denoting  the  spaces  between 
them  by  letters  placed  at  the  top,  and  bottom,  and  sides,  for 
readily  finding  any  field  or  other  object  mentioned  in  a  table. 

In  mapping  counties,  and  estates  that  have  uneven  grounds 
of  hills  and  valleys,  reduce  all  oblique  lines,  measured  up- 
hill and  down-hill,  to  horizontal  straight  lines,  if  that  was 
not  done  during  the  survey,  before  they  were  entered  in  the 
field-book,  by  making  a  proper  allowance  to  shorten  them.  For 
which  purpose  there  is  commonly  a  small  table  engraven  on 
some  of  the  instruments  for  surveying. 

THE  NEW  METHOD  OF  SURVEYING. 

PROBLEM  XV. 

To  Survey  and  Plan  by  the  JNTea:  Method. 

In  the  former  method  of  measuring  a  large  estate,  the  ac- 
curacy of  it  depends  both  on  the  correctness  of  the  instru- 
ments, and  on  the  care  in  taking  the  angles.  To  avoid  the 
errors  incident  to  such  a  multitude  of  angles,  other  methods 
have  of  late  years  been  used  by  some  few  skiful  surveyors  : 
the  most  practical,  expeditious,  and  correct,  seems  to  be  the 
following,  which  is  performed,  without  taking  angles,  by  mea- 
suring with  the  chain  only. 

Choose  two  or  more  eminences,  as  grand  stations,  and  mea- 
sure a  principal  base  Hne  from  one  station  to  another  ;  noting 
every  hedge,  brook,  or  other  remarkable  object,  as  you  pass 
by  it ;  measuring  also  such  short  perpendicular  lines  to  the 
bends  of  hedges  as  may  be  near  at  hand.  From  the  extre- 
mities of  this  base  line,  or  from  any  convenient  parts  of  the 
same,  go  off  with  other  lines  to  some  remarkable  object  situa- 
ted towards  the  sides  of  the  estate,  without  regarding  the 
angles  they  make  with  the  base  Ime  or  with  one  another  ; 
still  remembering  to  note  every  hedge,  brook,  or  other  object, 
that  you  pass  by.  These  lines,  when  laid  down  by  inter- 
sections, will  with  the  base  line,  form  a  grand  triangle  on 
the  estate  ;  several  of  which,  if  need  be,  being  thus  mea- 
sured and  laid  down,  you  may  proceed  to  form  other  smaller 
triangles  and  trapezoids  on  the  sides  of  the  former  j  and  so  on 
lill  you  finish  with  the  enclosures  individually.  By  which 
means  a  kind  of  skeleton  of  the  estate  may  first  be  obtained, 

and 


SURVEYING.  449 

and  the  chief  lines  serve  as  the  bases  of  such  triangles  and 
trapezoids  as  are  necessary  to  fill  up  all  the  interior  parts. 

The  field-book  is  ruled  into  three  columns,  as  usual.  In 
the  middle  one  are  set  down  the  distances  on  the  chain-line, 
at  which  any  mark,  offset,  or  other  observation,  is  made  ;  and 
in  the  right  and  left  hand  columns  are  entered  the  offsets  and 
observations  made  on  the  right  and  left  band  respectively  of 
the  chain-line  ;  sketching  on  the  sides  the  shape  or  resem- 
blance of  the  fences  or  boundaries. 

It  is  of  great  advantage,  both  for  brevity  and  perspicuity, 
to  begin  at  the  bottom  of  the  leaf,  and  write  upwards  ;  deno- 
ting the  crossing  of  fences,  by  lines  drawn  across  the  middle 
column,  or  only  a  part  of  such  a  line  on  the  right  and  left  op- 
posite the  figures,  to  avoid  confusion  ;  and  the  corners  of 
fields,  and  other  remarkable  turns  in  the  fences  where  offsets 
are  taken  to,  by  lines  joining  in  the  manner  the  fences  do  ;  as 
will  be  best  seen  by  comparing  the  book  with  the  plan  annex- 
ed to  the  field-book  following,  p.  450. 

The  letter  in  the  left-hand  corner  at  the  beginning  of  every 
line,  is  the  mark  or  place  measured /rom;  and  that  at  the 
right-hand  corner  at  the  end,  is  the  mark  measured  to :  But 
when  it  is  not  convenient  to  go  exactly  from  a  mark,  the 
place  measured  from  is  described  such  a  dUtanct  from  one. 
mark  towards  another  ;  and  where  a  former  mark  is  not  mea- 
sured to,  the  exact  place  is  ascertained  by  saying,  turn  to  the 
right  or  left  hand,  such  a  distance  to  such  a  mark^  it  being  al- 
ways understood  that  those  distances  are  taken  in  the  chain-line. 

The  characters  used  are,  f  for  turn  to  the  right  hand^^  for 
turn  to  the  left  hand,  and-^placed  over  an  offset,  to  show  that  it 
is  not  taken  at  right  angles  with  the  chain-line,  but  in  the  di- 
rection of  some  straight  fence  ;  being  chiefly  used  when  cross- 
ing their  directions  ;  which  is  a  better  way  of  obtaining  their 
true  places  than  by  offsets  at  right  angles. 

When  a  line  is  measured  whose  position  is  determined, 
either  by  former  work  (as  in  the  case  of  producing  a  giveo 
line,  or  measuring  from  one  known  place  or  mark  to  another) 
or  by  itself  (as  in  the  third  side  of  the  triangle),  it  is  called 
^fast  line,  and  a  double  line  across  the  book  is  drawn  at  the 
conclusion  of  it ;  but  if  its  position  is  not  determined  (as  in 
the  second  side  of  the  triangle),  it  is  called  a  loose  line,  and  a 
single  line  is  drawn  across  the  book.  When  a  line  becomes 
determined  in  position  and  is  afterwards  continued  fartheti 
a  double  line  half  through  the  book  is  drawn. 

When  a  loose  line  is  measured,  it  becomes  absolutely  ne- 
cessary to  measure  some  other  line  that  will  determine  its 
position.  Thus,  the  first  line  ah  or  hh^  being  the  base  of  a 
triangle  is  always  determined  ;  but  the  position  of  the  second 

Vol.  F.  58  side 


450  •  ,      LAND 

side  hj,  does  not  become  determined,  till  the  third  side  J6  is 
measured  ;  then  the  position  of  both  is  determined,  and  the 
triangle  may  be  constructed. 

At  the  beginning  of  a  line,  to  fix  a  loose  line  to  the  mark  or 
place  measured  from,  the  sign  of  turning  to  the  right  or  left 
hand  must  be  added,  as  at  h  in  the  second,  and  j  in  the  third 
line  ;  otherwise  a  strs^nger,  when  laying  down  the  work,  may 
as  easily  construct  the  triangle  hjb  on  the  wrong  side  of  the  line 
ahf  as  on  the  right  one  ;  but  this  error  cannot  be  fallen  into, 
if  the  sign  above  named  be  carefully  observed. 

In  choosing  a  line  to  fix  a  loose  one,  care  must  be  taken 
that  it  doe§  not  make  a  very  acute  or  obtuse  angle  ;  as  in  the 
t^^iangle  pBr,  by  the  angle  at  b  being  very  obtuse,  a  small  de- 
viation from  truth,  even  the  breadth  of  a  point  at  p  or  r, 
would  make  the  error  at  b,  when  constructed,  very  consider- 
able ;  but  by  constructing  the  triangle  pBg,  such  a  deviation  is 
of  no  consequence. 

Where  the  words  leave  off"  are  written  in  the  field-book,  it 
signifies  that  the  taking  of  ofisets  is  from  thence  discontinued  ; 
and  of  course  something  is  wanting  between  that  and  the  next 
offset,  to  be  afterwards  determined  by  measuring  gome  other 
line. 

The  field-book  for  this  method,  and  the  plan  drawn  froai 
it,  are  contained  in  the  four  following  pages,  engraven  on  cop- 
per-plates ;  answerable  to  which,  the  pupil  is  to  draw  a  plan 
from  the  measures  in  the  field-book,  of  a  larger  size,  viz.  to  a 
scale  of  a  double  size  will  be  convenient,  such  a  scale  being 
also  found  on  most  instruments.  In  doing  this,  begin  at  the 
commencement  of  the  field-book,  or  bottom  of  the  first  page 
and  draw  the  first,  line  ah  in  any  direction  at  pleasure,  and  then 
the  next  two  sides  of  the  first  triangle  bhj  by  sweeping  inter- 
secting arcs  ;  and  so  all  the  triangles  in  the  same  manner,  after 
each  other  in  their  order  ;  and  afterwards  setting  the  perpendi- 
cular and  other  offsets  at  their  proper  places,  and  through  the 
ends' of  them  drawing  the  bounding  fences. 

J^ote.  That  the  field-book  begins  at  the  bottom  of  the  first 
page,  and  reads  up  to  the  top  ;  hence  it  goes  to  the  bottom  of 
the  next  page,  and  to  the  top  ;  and  thence  it  passes  from  the 
bottom  of  the  third  page  to  the  top  which  is  the  end  of  the 
field-book.  The  several  marks  measured  to  or  from,  are  here 
denoted  by  the  letters  of  the  alphabet,  first  the  small  ones 
«,  6,  c,  df  &c,  and  after  them  the  capitals  A^  B,  C,  Z>,  he.  But, 
instead  of  these  letters,  some  surveyors  use  thq  numbers  in 
order,  1,  2,  3,  4,  &c. 


OF 


SURVEYING. 


45} 


OP  THB  OLD  KIND  OF  FIELD-BOOK. 

In  surveying  with  the  plain  table,  a.  field-book  is  not  used, 
as  every  thing  is  drawn  on  the  table  immediately  when  it  is 
measured.  But  in  surveying  with  the  theodolite,  or  any  other 
instrument,  some  kind  of  a  field  book  must  be  used,  to  write 
down  in  it  a  register  or  account  of  all  that  is  done  and  occurs 
relative  to  the  survey  in  hand. 

This  book  every  one  contrives  and  rules  as  he  thinks  fittest 
for  himself.  The  following  is  a  specimen  of  a  form  which  has 
been   formerly  used.     It  is  ruled  into  three  columns,  as  below. 

Here  ®  1  is  the  first  station,  where  the  angle  or  bearing  is 
10r)*»  25'.  On  the  left,  at  73  links  in  the  distance  or  principal 
line,  is  an  ofi*set  of  92  ;  and  at  610  an  offset  of  24  to  a  cross 
hedge.  On  the  right  at  0,  or  the  beginning,  an  offset  26  to 
the  corner  of  the  field  ;  at  248  Brown's  boundary  hedge  com- 
mences ;  at  610  an  offset  35  ;  and  at  954,  the  end  of  the  first 
line,  the  0  denotes  its  terminating  in  the  hedge.  And  so  on 
for  the  other  stations- 

A  line  is  drawn  under  the  work,  at  the  end  of  every  station 
line,  to  prevent  confusion. 

Form  of  this  Field-Book. 


Stations, 

Offsets  and  Remarks 

Bearings, 

Offsets  and  Remarks 

on  the  left. 

and 

on  the  right. 

Distances. 

®    1 

105°  25' 

80 

00 

25  corner 

92 

.     73 

248 

Brown's  hedge 

a  cross  hedge  24 

610 

35 

954 

00 

®    2 

53«»   10' 

house  corner  51 

23 

21 

120 

29  a  tree 

34 

734 

40  a  stile 

0    3 

670  20' 

61 

35 

a  brook          30 

248 

639 

16  a  spring 

foot  path          16 

810 

<:ross  hedge     18 

973 

20  a  pond           Th( 

45t 


tANB 


Then  the  plan,  ob  a  small  scale  drawn  from  the  above  field- 
book,  will  be  as  in  the  following  figure.  But  the  pupil  may 
draw  a  plan  of  3  or  4  times  the  size  on  his  paper  book.  The 
dotted  lines  denote  the  3  chain  or  measured  lines,  and  the 
black  lines  the  boundaries  on  the  right  and  left. 


But  some  skilful  surveyors  now  make  use  of  a  different 
method  for  the  field-book,  namely,  beginning  at  the  bottom 
of  the  page  and  writing  upwards  ;  sketching  also  a  neat 
boundary  on  either  hand  resembling  the  parts  near  the 
measured  lines  as  they  pass  along  ;  an  example  of  which  will 
be  given  further  on,  in  the  method  of  surveying  a  large  estate. 

In  smaller  surveys  and  measurements,  a  good  way  of  setting 
down  the  work,  is,  to  draw  by  the  eye  on  a  piece  of  paper,  a 
a  figure  resembling  that  which  is  to  be  measured  ;  and  so 
writing  the  dimensions,  as  they  are  found,  against  the  corres- 
ponding parts  of  the  figure.  And  this  method  may  be  practis- 
ed to  a  considerable  extent,  even  in  the  larger  surveys. 

Another  specimen  of  a  field-book,  with  its  plan,  is  as  fol- 
lows ;  being  a  single  field,  surveyed  with  the  chain,  and  the 
theodolite  for  taking  angles  ;  which  the  pupil  will  likewise 
draw  of  a  larger  size. 


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SURVEYING.  453 


SECTION  III. 

OF  COMPUTING  AND  DIVIDING. 

PROBLEM  XVI. 

To  Compute  the  Contents  of  Fields, 

1.  Compute  the  contents  of  the  figures  as  divided  into 
triangles,  or  trapeziums,  by  the  proper  rules  for  these  figures 
laid  down  in  measuring  ;  multiplying  the  perpendiculars  by 
the  diagonals  or  bases,  both  in  links,  and  divide  by  2  ;  the 
quotient  is  acres,  after  having  cut  oflf  five  figures  on  the  right 
for  decimals.  Then  bring  these  decimals  to  roods  and  perches, 
by  multiplying  first  by  4,  and  then  by  40.  An  example  of 
which  is  given  in  the  description  of  the  chain,  pag.  429. 

2.  In  small  and  separate  pieces,  it  is  usual  to  compute  their 
contents  from  the  measures  of  the  lines  taken  in  surveying 
them,  without  making  a  correct  plan  of  them. 

3.  In  pieces  bounded  by  very  crooked  and  winding  hedges, 
measured  by  offsets,  all  the  parts  between  the  offsets  are  most 
accurately  measured  separately  as  small  trapezoids. 

4.  Sometimes  such  pieces  as  that  last  mentioned,  are  com- 
puted by  finding  a  mean  breadth,  by  adding  all  the  offsets 
together,  and  dividing  the  sum  by  the  number  of  them,  ac- 
counting that  for  one  of  them  where  the  boundary  meets 
the  station-line,  (which  increases  the  number  of  them  by  1, 
for  the  divisor,  though  it  does  not  increase  the  sum  or 
quantity  to  be  divided)  ;  then  multiply  the  length  by  that 
mean  breadth. 

5.  But  in  larger  pieces  and  whole  estates,  coosistiDg  of 
many  fields,  it  is  the  common  practice  to  make  a  rough  plan 
of  the  whole,  and  from  it  compute  the  contents,  quite  inde- 
pendent of  the  measures  of  the  lines  and  angles  that  were 
taken  in  surveying.    For  then  new  lines  are  drawn  in  the 

fieldi! 


454  LAND 

fields  on  the  plan,  so  as  to  divide  them  into  trapeziums  and 
triangles,  the  bases  and  perpendiculars  of  which  are  mea- 
sured on  the  plan  by  means  of  the  scale  from  which  it  was 
drawn,  and  so  multiplied  together  for  the  contents.  In  this 
way,  the  work  is  very  expeditiously  done,  and  sufficiently 
correct  ;  for  such  dimensions  are  taken  as  afford  the  most 
easy  method  of  calculation  ;  and  among  a  number  of  parts, 
thus  taken  and  applied  to  a  scale,  though  it  be  likely  that 
some  of  the  parts  will  be  taken  a  small  matter  too  little,  and 
others  too  great,  yet  they  will,  on  the  whole,  in  all  probability, 
very  nearly  balance  one  another,  and  give  a  sufficiently  ac- 
curate result.  After  all  the  fields  and  particular  parts  are 
thus  computed  separately,  and  added  all  together  into  one 
sum  ;  calculate  the  whole  estate  independent  of  the  fields,  by 
dividing  it  into  large  and  arbitrary  triangles  and  trapeziums, 
and  add  these  also  together.  Then  if  this  sum  be  equal  to 
the  former,  or  nearly  so,  the  work  is  right  ;  but  if  the  sums 
have  any  considerable  difference,  it  is  wrong,  and  they 
must  be  examined,  and  re-computed,  till  they  nearly  agree. 
.  6.  But  the  chief  art  in  computing,  consists  in  finding 
the  contents  of  pieces  bounded  by  curved  or  very  irregular 
lines,  or  in  reducing  such  crooked  sides  of  fields  or  boun- 
daries to  straight  lines,  that  shall  inclose  the  same  or  equal 
area  with  those  crooked  sides,  and  so  obtain  the  area  of  the 
curved  figure  by  means  of  the  right-lined  one,  which  will 
commonly  be  a  trapezium.  Now  this  reducing  the  crooked 
sides  to  straight  ones,  is  very  easily  and  accurately  performed 
in  this  manner  : — Apply  the  straight  edge  of  a  thin,  clear 
piece  of  lanthorn-horn  to  the  crooked  line,  which  is  to  be 
reduced,  in  such  a  manner,  that  the  small  parts  cut  off  from 
the  crooked  figure  by  it,  may  be  equal  to  those  which  are 
taken  in  :  which  equality  of  the  parts  included  and  excluded 
you  will  presently  be  able  to  judge  of  very  nicely  by  a  little 
practice  ;  then  with  a  pencil,  or  point  of  a  tracer,  draw  a 
line  by  the  straight  edge  of  the  horn.  Do  the  same  by  the 
other  sides  of  the  field  or  figure.  So  shall  you  have  a  straight- 
sided  figure  equal  to  the  curved  one  ;  the  content  of  which, 
being  computed  as  before  directed,  will  be  the  content  of  the 
crooked  figure  proposed. 

Or,  instead  of  the  straight  edge  of  the  horn,  a  horse  hair, 
or  fine  thread,  may  be  applied  across  the  crooked  sides  in  the 
same  manner  ;  and  the  easiest  way  of  using  the  thread,  is  to 
string  a  small  slender  bow  with  it,  either  of  wire  or  cane,  or 
whale-bone,  or  such-Hke  slender  elastic  matter  ;  for  the  bow 
keeping  it  always  stretched,  it  can  be  easily  and  neatly  ap- 
plied with  one  hand,  while  the  other  is  at  liberty  to  make 
two  marks  by  the  side  of  it,  to  draw  the  straight  line  by. 

EXAMPLE. 


SURVEYING,  4^5 

EXAMPLE. 

Thus,  let  it  be  required  to  find  the  contents  of  the  same 
figure  as  ia  Prob.  ix,  page  4U,  to  a  scale  of  4  chains  to  aa 
inch. 


A 


B 


C 

Draw  the  4  dotted  straight  lines  ab,  bc,  cd,  da,  cutting  ofl" 
equal  quantities  on  both  sides  of  them,  which  they  do  as  near 
as  the  eye  can  judge  :  so  is  the  crooked  figure  reduced  to  an 
equivalent  right-lined  one  of  4  sides  abcd.  Then  draw  the 
diagonal  bd,  which,  by  applying  a  proper  scale  to  it,  measures 
suppose  1256.  Also  the  perpendicular,  or  nearest  distance 
from  A  to  this  diagonal,  measures  456  ;  and  the  distance  of  c 
from  it,  is  428. 

Then,  half  the  sum  of  456  and  428,  multiplied  by  the  dia- 
gonal 1256,  gives  555152  square  links,  or  5  acres,  2  roods, 
8  perches,  the  content  of  the  trapezium,  or  of  the  irregular 
crooked  piece. 

Asa  general  example  of  this  practice,  let  the  contents  be 
computed  of  all  the  fields  separately  in  the  foregoing  plan  ia 
page  452,  and  by  adding  the  contents  altogether,  the  whole 
sum  or  content  of  the  estate  will  be  found  nearly  equal  to 
103|  acres.  Then,  to  prove  the  work,  divide  the  whole  plan 
into  two  parts,  b}^  a  pencil  line  drawn  across  it  any  way  near 
the  middle,  as  from  the  corner  I  on  the  right,  to  the  corner 
near  s  on  the  left ;  then  by  computing  these  two  large  parts 
separately,  their  sum  must  be  nearly  equal  to  the  former  sura, 
when  the  work  is  all  right. 

PROBLEM  XVU. 

To  Transfer  a  Plan  to  Another  Paper ^  ^c. 

After  the  rough  plan  is  completed,  and  a  fair  one  is  want- 
ed ;  tkis  may  be  done  by  any  of  the  following  methods. 

First 


456  LAND 

First  Method, — Lay  the  rough  plan  on  the  clean  papei 
keeping  them  always  pressed  flat  and  close  together,  by 
"  weights  laid  on  them.  Then,  with  the  point  of  a  fine  pin  oi 
pricker,  prick  through  all  the  corners  of  the  plan  to  be  copi- 
ed. Take  them  asunder,  and  connect  the  pricked  points,  on 
the  clean  paper,  with  lines  ;  and  it  is  done.  This  method 
is  only  to  be  practised  in  plans  of  such  figures  as  are  small 
and  tolerably  regular,  or  bounded  by  right  lines. 

Second  Method. — Rub  the  back  of  the  rough  plan  over  with 
black-lead  powder ;  and  lay  this  blacked  part  on  the  cleaa 
paper  on  which  the  plan  is  to  be  copied,  and  in  the  proper 
position.  Then,  with  the  blunt  point  of  some  hard  substance, 
as  brass  or  such-like,  trace  over  the  lines  of  the  whole  plan  ; 
pressing  the  tracer  so  much,  as  that  the  black  lead  under  the 
lines  may  be  transferred  to  the  clean  paper :  after  which, 
take  oflT  the  rough  plan,  and  trace  over  the  leaden  marks 
with  common  ink,  or  with  Indian  ink — Or,  instead  of  blacking 
the  rough  plan,  we  may  keep  constantly  a  blacked  paper  to 
lay  between  the  plans. 

Third  Method. — Another  method  of  copying  plans,  is  by 
means  of  squares.  This  is  performed  by  dividing  both  ends 
and  sides  of  the  plan  which  is  to  be  copied  into  any  conve- 
nient number  of  equal  parts,  and  connecting  the  correspond- 
ing points  of  division  with  lines  :  which  will  divide  the  plan 
into  a  number  of  small  squares.  Then  divide  the  paper,  on 
which  the  plan  is  to  be  copied,  into  the  same  number  of 
squares,  each  equal  to  the  former  when  the  plan  is  to  be  co- 
pied of  the  same  size,  but  greater  or  less  than  the  others,  in 
the  proportion  in  which  the  plan  is  to  be  increased  or  dimin- 
ished, when  of  a  different  size.  Lastly,  copy  into  the  clean 
squares  the  parts  contained  in  the  corresponding  squares  of 
the  old  plan  ;  and  you  will  have  the  copy,  either  of  the  same 
size,  or  greater  or  less  in  any  proportion. 

Fourth  Method. — A  fourth  method  is  by  the  instrument  call- 
ed a  pentagraph,  which  also  copies  the  plan  in  any  size  re- 
quired. 

Fifth  Method. — But  the  neatest  method  of  any,  at  least  in 
copying  from  a  fair  plan,  is  this.  Procure  a  copying  frame 
or  glass,  made  in  this  manner  ;  namely,  a  large  square  of  the 
best  window  glass,  set  in  a  broad  frame  of  wood,  which  can  be 
raised  up  to  any  angle,  when  the  lower  side  of  it  rests  on  a 
table.  Set  this  frame  up  to  any  angle  before  you,  facing  a 
strong  light ;  fix  the  old  plan  and  clean  paper  together,  with 
several  pins  quite  around,  to  keep  them  together,  the  clean 

paper 


SURVEYING.  457 

paper  being  laid  uppermost,  and  over  the  face  of  the  plan  to 
be  copied  Lay  them,  with  the  back  of  the  old  plan,  on  the 
glass  ;  namely,  that  part  which  you  intend  to  bejj;in  at  to  copy 
first  ;  and  by  means  of  the  light  shining  through  the  papers 
you  will  very  distinctly  perceive  every  line  of  the  plan 
through  the  clean  paper.  In  this  state  then  trace  all  the  lines 
on  the  paper  with  a  pencil  Having  drawn  that  part  which 
covers  the  glass,  slide  another  part  over  the  glass,  and  copy  it 
in  the  same  manner.  Then  another  part.  And  so  on,  till 
"the  whole  is  copied.  Then  take  them  asunder,  and  trace 
all  the  pencil  lines  over  with  a  ^ne  pen  and  Indian  ink,  or 
with  common  ink.  And  thus  you  may  copy  the  finest  plan 
without  injuring  it  in  the  least. 


OF  ARTIFICERS'  WORKS, 

AND 

TIMBER  MEASURING. 


I.  OF  THE  CARPENTER'S  OR  SLIDING  RULE. 

THE  Carpenter's  or  Sliding  Rule,  is  an  instrument  much 
tised  in  measuring  of  timber  and  artificers'  works,  both  for 
taking  the  dimensions,  and  computing  the  contents. 

The  instrument  consists  of  two  equal  pieces,  each  a  foot  in 
length,  which  are  connected  together  by  a  folding  joint.  * 

One  side  or  face  of  the  rule,  is  divided  into  inches,  and 
eighths,  or  half-quarters.  On  the  same  face  also  are  several 
plane  scales,  divided  into  twelfth  parts  by  diagonal  lines  ; 
which  are  used  in  planning  dimensions  that  are  taken  in  feet 
and  inches.  The  edge  of  the  rule  is  commonly  divided  de- 
cimally, or  into  tenths  ;  namely,  each  foot  into  ten  equal 
parts,  and  each  of  these  into  ten  parts  again  :  so  that  by 
me£lns  of  this  last  scale,  dimensions  are  taken  in  feet,  tenths, 
and  hundredths,  and  multiplied  as  common  decimal  numbers, 
which  is  the  best  way. 

On  the  one  part  of  the  other  face  are  four  lines,  marked 
A,  B,  c,  D  ;  the  two  middle  ones  b  and  c  being  on  a  slider, 
which  runs  in  a  groove  made  in  the  stock.  The  same  num- 
bers serve  for  both  these  two  middle  lines,  the  one  being 
above  the  numbers,  and  the  other  below. 

Vol.  i.  59  These. 


458  BRICKLAYERS'  WORK. 

These  fonr  lines  are  logarithmic  ones,  and  the  three  a,  h, 
c,  which  are  all  equal  to  one  another,  are  double  lines,  as 
they  proceed  twice  over  from  1  to  10.  The  other  or  lowest 
line,  D,  is  a  single  one,  proceeding  from  4  to  40.  It  is  also 
called  the  girt  line,  from  its  use  in  computing  the  contents 
of  trees  and  timber  ;  and  on  it  are  marked  wg  at  17- 15,  and 
AG  at  18'95,  the  wine  and  ale  gage  points,  to  make  this  in*- 
strument  serve  the  purpose  of  a  gaging  rule. 

On  the  other  part  of  this  face,  there  is  a  table  of  the  value 
of  a  load,  or  30  cubic  feet,  of  timber,  at  all  prices,  from  6 
pence  to  2  shillings  a  foot. 

When  1  at  the  beginning  of  any  line  is  accounted  1,  then 
the  1  in  the  middle  will  be  10,  and. the  10  at  the  end  100  ; 
but  when  1  at  the  beginning  is  counted  10,  then  the  1  in 
the  middle  is  100,  and  the  ^0  at  the  end  ioOO  ;  and  so  on. 
And  all  the  smaller  divisions  are  altered  proportionally. 


II.  ARTIFICERS'  WORK. 

Artificers  compute  the  contents  of  their  works  by  several 
different  measures.     As, 

Glazing  and  masonry,  by  the  foot  ;  Painting,  plastering, 
paving,  &c.  by  the  yard,  of  9  square  feet  :  Flooring, 
partitioning,  roofing,  tiling,  &c.  by  the  square  of  100 
square  feet  : 

And  brickwork,  either  by  the  yard  of  9  square  feet,  or  by 
the  perch,  or  square  rod  or  pole,  containing  212\  square 
feet,  or  30|  square  yards,  being  the  square  of  the  rod 
or  pole  of  161  feet  or  5^  yards  long. 

As  this  number  ?72i  is  troublesome  to  divide  by,  the  ^  is 
of\en  omitted  in  practice,  and  the  content  in  feet  divided  only 
by  the  272. 

All  works  whether  superficial  or  solid,  are  computed  by 
the  rules  proper  to  the  figure  of  them,  whether  it  be  a  triangle, 
or  rectangle^  a  parallelopiped,  or  any  other  figure. 


III.  BRICKLAYERS'  WORK. 

Brickwork  is  estimated  at  the  rate  of  a  brick  and  a  half 
thick,  bo  that  if  a  wall  be  more  or  less  than  this  standard 
thickness,  it  must  be  reduced  to  it,  as  follows  : 

Multiply  the  superficial  content  of  the  Wall  by  the  number 
of  half  bricks  in  the  thickness,  and  divide  the  product  by  3. 

The 


MASONS'  WORK.  469 

The  dimensions  of  a  building  may  be  taken  by  measuring 
half  round  on  the  outside  and  half  round  it  on  the  inside  ;  the 
sum  of  these  two  gives  the  compass  of  the  wall,  to  be  multi- 
plied by  the  height,  for  the  content  of  the  materials. 

Chimneys  are  commonly  measured  as  if  they  were  solid, 
deducting  only  the  vacuity  from  the  hearth  to  the  mantle, 
oq  account  of  the  troubl<»  of  them.  All  windows,  doors,  &c. 
are  to  be  deducted  out  of  the  contents  of  the  walls  in  which 
they  are  placed. 

EXAMPLES. 

Exam.  1 .  Hour  many  yards  and  rods  of  standard  brick-work 
are  in  a  wall  whose  length  or  compass  is  57  feet  3  inches,  and 
height  24  feet  6  inches  ;  the  wall  being  2^  bricks  or  5  half- 
brirks  thick  ?  Ans.  8  rods,  17f  yards. 

Exam.  2.  Required  the,  content  of  a  wall  62  feet  6  inches 
long,  and  14  feet  8  inches  high,  and  2^  bricks  thick  ? 

Aiis.  169-763  yards. 

Exam.  3.  A  triangular  gable  is  raised  17i  feet  high,  on 
an  end  wall  whose  length  is  24  feet  9  inches,  the  thickness 
being  2  bricks  :  required  the  reduced  content  ? 

Ans.  32-08^  yards. 

Exam.  4.  The  end  wall  of  a  house  is  28  feet  10  inches 
long,  and  nb  feet  8  inches  high,  to  the  eaves  ;  20  feet  high  is 
Q\  bricks  thick,  other  20  feet  high  is  2  bricks  thick,  and  the 
remaining  15  feet  8  inches  is  H  brick  thick  ;  above  which  is 
a  triangular  gaSle,  of  1  brick  thi<k  ;  which  rises  42  courses  of 
bricks,  of  which  every  4  courses  make  a  foot.  What  is  the 
whole  content  in  standard  measure  ?  Ans.  253626  yards. 


IV.  MASONS'  WORK. 

To  masonry  belong  all  sorts  of  stone-work  ;  and  the  mea- 
sure made  use  of  is  a  foot,  either  superficial  or  solid. 

Walls,  columns,  blocks  of  j^tone  or  marble,  &c.  are  mea- 
sured by  the  cubic  foot ;  and  pavements,  slabs,  chimney-pieces, 
&c.  by  the  sui)erficial  or  square  foot. 

Cubic  or  solid  measure  is  used  for  the  materials,  and  square 
measure  for  the  workmanship. 

In  the  sohd  measure,  the  true  length,  breadth,  and  thick- 
ness are  taken  and  multiplied  continually  together.  In  the 
superticial,  there  must  be  taken  the  length  and  breaith  of 
every  part  of  the  projection  which  is  seen  without  the  gene- 
ral upright  face  of  the  building. 

EXAMPUES. 


46©        CARPENTERS'  AND  JOINERS'  WORK. 
EXAMPLES. 

Exam.  1.  Required  the  solid  content  of  a  wall,  63  feet  tf 
inches  long,  12  feet  3  inches  high,  and  2  feet  thick  ? 

Ans.  1310f  feet. 

Exam.  2.  What  is  the  solid  content  of  a  wall,  the  length 
being  24  feet  3  inches,  height  10  feet  9  inches,  and  2  feet 
thick?  Ans.  621-375  feet. 

Exam.  3.  Required  the  value  of  a  marble  slab,  at  8s.  per 
foot  ;  the  length  being  6  feet  7  inches,  and  breadth  1  foot  10 
inches  ?  Ans.  4/.  Is.  lO^d, 

Exam,  4-   In  a  chimney-piece,  suppose  the 
length  of  the  mantle  and  slab,  each  4  feet  6  inches 
breadth  of  both  together        ;-         3         2 
length  of  each  jamb        -         -         4         4 
breadth  of  both  together         -  1  9 

Required  the  superficial  content  ?  Ans.  21  feet  10  inches. 


V.  CARPENTERS'  AND  JOINERS'  WORK, 

To  this  branch  belongs  all  the  wood-work  of  a  house,  such 
as  flooring,  partitioning,  rooting,  &c. 

Large  and  plain  articles  are  usually  measured  by  the 
square  foot  or  yard,  &.c.  ;  but  enriched  mouldings,  and  ^ome 
other  articles,  are  often  estimated  by  running  or  iineal  mea- 
sure ;  and  some  things  are  rated  by  the  piece. 

la  measuring  of  Joists,  take   the  dimensions  of  one  joist, 
and  multiply  its  content  by  the  number  of  them  ;    considtMing  ■ 
that  each  end  is   let  into  the  wall   about  |  of  the  thickness, 
as  it  ought  to  be. 

Partitions  are  measured  from  wall  to  wall  for  one  dimen- 
sion, and  from  floor  to  floor,  as  far  as  they  extend,  for  the 
other. 

The  measure  of  Centering  for  Cellars  is  foiind  by  making  a 
string  pass  over  the  surface  of  the  arch  for  the  breadth,  and 
taking  the  length  of  the  cellar  for  the  length  :  but  in  groin 
centering,  it  is  usual  to  allow  double  measure,  on  account  of 
their  extraordinary  trouble. 

In  Roofing,  the  dimensions  as  to  length,  breadth,  and  depth, 
are  taken  as  in  flooring  joists,  and  the  contents  computed  the 
same  way. 

In  Floor-boarding,  take  the  length  of  the  room  for  one  di- 
mension, and  the  breadth  for  the  other,  to  multiply  together 
for  the  content. 

For  Stair-cases,  take  the  breadth  of  all  the  steps,  by  ma>nng 

a  line 


CARPENTERS'  AND  JOINERS'  WORk.         461 

a  line  ply  close  over  them,  from  the  top  to  the  bottom,  and 
multiply  the  length  of  this  line  by  the  length  of  a  step,  for 
the  whole  area. — By  the  length  of  a  step  is  meant  the  length 
of  the  front  and  the  returns  at  the  two  ends  ;  and  by  the 
breadth  is  to  be  understood  the  girts  of  its  two  outer  surfaces, 
er  the  tread  and  riser. 

For  the  Balustrade^  take  the  whole  length  of  the  upper  part 
of  the  hand-rail,  and  girt  over  its  end  till  it  meet  the  top  of 
the  newel  post,  for  the  one  dimension  ;  and  twice  the  length 
of  the  baluster  on  the  landing,  with  the  girt  of  the  hand-rail, 
for  the  other  dimension. 

For  Wainscoting^  take  the  compass  of«  the  room  for  the 
one  dimension  ;  and  the  height  from  the  floor  to  the  ceiling, 
making  the  string  ply  close  into  all  the  mouldings,  for  the 
other. 

For  Doors y  take  the  height  and  the  breadth,  to  multiply 
them  together  for  the  area. — If  the  door  be  panneled  on  both 
sides,  take  double  its  measure  for  the  workmanship  ;  but 
if  one  side  only  be  panneled,  take  the  area  and  its  half  for 
the  workmanship. — For  the  Surrounding  Architrave^  girt  it 
about  the  uppermost  part  for  its  length  ;  and  measure  over  it, 
as  far  as  it  can  be  seen  when  the  door  is  open,  for  the 
breadth. 

Window-shutters ^  Bases,  &c.  are  measured   in  like  manner. 

In  measuring  of  Joiners'  work,  the  string  is  made  to  ply 
close  into  all  the  mouldings,  and  to  every  part  of  the  work 
over  which  it  passes. 

EXAMPLES. 

Exam.  1.  Required  the  content  of  a  floor,  48  feet  6  inches 
long,  and  24  feet  3  inches  broad  ?  Ans.   1 1  sq.  16}  feet. 

Exam.  2.  A  floor  being  36  feet  3  inches  long,  and  16  feet 
6  inches  broad,  how  many  squares  are  in  it  ? 

Ans.  5  sq.  98i  feet. 

ExAivr.  3.  How  many  squares  are  there  in  173  feet  10 
inches  in  length,  ^and  10  feet  7  inches  height,  of  partition- 
ing ?  Ans.  18-3973  squares. 

BxAM.  4.  What  cost  the  roofing  of  a  house  at  10s.  6d. 
a  square  ;  the  length  within  the  walls  being  62  feet  8  inches, 
and  the  breadth  30  feet  6  inches  ;  reckoning  the  roof  f  of 
the  flat  ?  Ans.  121,  ns.  life/. 

Exam. 


462  SLATERS*  AND  TILERS'  WORK. 

Exam.  5.  To  how  much,  at  6s.  per  square  jard,  amounte 
the  wainscoting  of  a  room  ;  the  height,  taking  in  the  cor- 
nice and  mouldings,  being  12  feet  6  inches,  and  the  whole 
compass  83  feet  8  inches  ;  also  the  three  window- shutters 
are  each  7  feet  8  inches  by  3  feet  6  inches,  and  the  door 
7  feet  by  3  feet  6  inches  ;  the  doors  and  shutlers,  being 
worked  on  both  sides,  are  reckoned  work  and  half  work  ? 

Aas.  36/.  12s.  2^d. 


VI.  SLATERS'  AND  TILERS'  WORK. 

In  these  articles,  the  content  of  a  roof  is  found  by  multi- 
plying the  length  of  the  ridge  by  the  girt  over  from  eaves  to 
eaves  ;  making  allowance  in  this  girt  for  the  double  row  of 
slates  at  the  bottom,  or  for  how  much  one  row  of  slates  or 
tiles  is  laid  over  another. 

When  the  roof  is  of  a  true  pitch,  that  is,  forming  a  right 
angle  at  top  ;  then  the  breadth  of  the  building,  with  its  half 
added,  is  the  girt  over  both  sides  nearly. 

In  angles  formed  in  a  roof,   running  from  the  ridge  to  the 
eaves,  when  the  angle   bends  inwards,  it  is  called  a  valley  j 
but  when  outwards,  it  is  called  a  hip- 
Deductions  are  made  for  chimney  shafts  or  window  holes. 

EXAMPLES. 

Exam.  1.  Required  the  content  of  a  slated  roof,  the  length 
being  45  feet  9  inches,  and  the  whole  girt  34  feet  3  inches  ? 

Ans.    174j^g  yards. 

Exam.  2.  To  how  much  amounts  the  tihng  of  a  house, 
at  2)5.  6d.  per  square  ;  the  length  being  43  feet  10  inches, 
and  the  breadth  on  the  flat  27  feet  6  inches  ;  also  the  eaves 
projecting  16  inches  on  each  side,  and  the  roof  of  a  true 
pitch  ?  Ans.  24^.  9s.  6-3</. 


VII.  PLASTERERS'  WORK. 

Plasterers'  work  is  of  two  kinds  ;  namely,  ceiling,  which 
is  plastering  on  laths  ;  and  rendering,  which  is  plastering  on 
vifails  :  which  are  measured  separately. 

Th^ 


PAINTERS'  WORK.  466 

The  contents  are  estimated  either  by  the  foot  or  the  yard, 
or  the  sfjuare»  of  100  feet.  Enriched  mouldings,  Lc.  are  rate* 
by  mulling  or  lineal  measure. 

Deductions  are  made  for  chimneys,  doors,  windows,  &c. 

EXAMPLES. 

Exam,  1.  How  many  yards  contains  the  ceiling  which  is  43 
feet  3  inches  long,  and  26  feet  6  inches  broad  ? 

Ans.  122i. 

Exam.  2.  To  how  much  amounts  the  ceiling  of  a  room,  at 
10^.  per  yard  ;  the  lensfth  being  21  feet  8  inches,  and  the 
breadth  !4  feet  10  inches.  Ans.  11.  9s.  8fc?, 

Exam.  3.  The  leni^th  of  a  room  is  18  feet  6  inches,  the 
breadth  12  feet  3  inches,  and  heis^ht  10  feet  6  inches  ;  to  how 
much  amounts  the  ceiling  and  rendering,  the  former  at  8d  and 
the  latter  at  Sd.  per  yard  ;  allowing  for  the  door  of  7  feet  by 
3  feet  8,  and  a  fire-place  of  5  feet  square  ? 

Ans.  1/.  13s.  3^d, 
Exam.  4.  Required  the  quantity  of  plastering  in  a  room, 
the  length  being  14  feet  5  inches,  breadth  13  feet  2  inches, 
and  height  9  feet  3  inches  to  the  under  side  of  the  cornice, 
which  girts  8i  inches,  and  projects  5  inches  from  the  wall  on 
the  upper  part  next  the  ceiling  ;  deducting  only  for  a  door  7 
feet  by  4  ? 

Ans.  63  yards  6  feet  3^  inches  of  rendering 
18  6  6  of  ceihng 

39  Of  i  of  cornice. 


VIII.  PAINTERS'  WORK. 

Painters'  work  is  computed  in  square  yards.  Every  part 
is  measured  where  the  colour  lies  ;  and  the  measuring  line  is 
forced  into  all  the  mouldings  and  corners. 

Wiiviows  are  done  at  so  much  a  piece.  And  it  is  usual  t© 
allow  double  measure  for  carved  mouldings,  &c. 

EXAMPLES. 

Exam.  1.  How  many  yards  of  painting  contains  the  room 
which  is  65  feet  6  inches  in  compass,  and  12  feet  4  inches 
high  ?  .    Ans.  89fi  yards. 

Exam.  2,  The  length  of  a  room  being  20  feet,  its  breadth 

14  feet 


464  GLAZIERS'  WORK. 

H  feet  6  inches,  and  height  10  feet  4  inches  ;  how  many 
yards  of  painting  are  in  it,  deducting  a  fire-place  of  4  feet  by 
4  feet  4  inches,  and  two  windows  each  6  feet  by  3  feet  2  in- 
ches ?  An&.  IS-^j  yards. 
Exam.  3.  What  cost  the  painting  of  a  room,  at  6d.  per 
yard  ;  its  length  being  24  feet  6  inphes,  its  breadth  16  feet  .3 
inches,  and  height  12  feet  9  inches  ;  also  the  door  is  7  feet  by 
3  feet  6,  and  the  window-shutters  to  two  windows  each  7  feet  9 
by  3  feet  6  ;  but  the  breaks  of  the  windows  themselves  are  8 
feet  6  inches  high,  and  1  foot  3  inches  deep  ;  including  also 
the  window  sills  or  seats,  and  the  soffits  above,  the  dimensions 
of  which  are  known  from  the  other  dimensions  :  but  deduct- 
ing the  fire-place  of  6  feet  by  5  feet  6  ? 

Ans.  3/.  3s.  lO^d- 


IX.  GLAZIERS'  WORK. 

Glaziers  take  their  dimensions,  either  in  feet,  inches  and 
parts,  or  feet,  tenths,  and  hundredths.  And  they  compute 
their  work  in  square  feet. 

In  taking  the  length  and  breadth  of  a  window,  the  cross 
l»ars  between  the  squares  are  included.  Also  windows  of 
round  or  oval  forms  are  measured  as  square,  measuring  them 
to  their  greatest  length  and  breadth,  on  account  of  the  waste 
in  cutting  the  glass. 

EXAMPLES. 

Exam.   1.  How  many  square  feet  contains  th6  window  which 

is  4-25  feet  long,  and  2-75  feet  broad  ?  Ans.  1  If. 

Exam.  2.  What  will  the  glazing  a  triangular  sky -light  come 

to,  at  lOd.  per  foot ;  the  base  being  12  feet  6  inches,  and  the 

perpendicular  height  6  feet  9  inches  ? 

Ans.  11.  15s.  Ifd. 
Exam.  3.  Ther^  is  a  house  with  three  tiers  of  windows, 
three  windows  in  each  tier,  their  common  breadth  3  fe€t  1 1 
inches  : 

now  the  height  of  the  first  tier  is  7  feet  10  inches 
of  the  second       6  8 

of  the  third  5  4 

Required  the  expense  of  glazing  at  I4d.  per  foot  ? 

Ans.  13/.  lis.  lO^d. 
Exam. 


PLUMBERS'  WORK.  465 

RxAM.  4.  Required  the  expense  of  glazing;  the  windows 
ef  a  hoase  at  13i.^a  foot  ;  there  being  three  stories,  and  three 
windows  in  each  storjr  : 

the  height  of  the  lower  tier  is  7  feet  9  inches 
of  the  middie  6  6 

of  the  upper  5  3^ 

and  of  an  oval  window  over  the  door  1         10^ 
the  common  breadth  of  ail  the  windows  being  3  feet  9  inches  ? 

Ans,  12L  OS,  6d, 


X.  PAVERS'  WORK. 

Pavers'  work  is  done  by  the  square  yard.     And  the  con- 
tent is  found  by  multiplying  the  length  by  the  breadth.         ' 


EXAMPLES. 

Exam.  1.  What  cost  the  paving  a  foot-jpath,  at  3s.  4c?, 
a  yard  ;  the  length  being  35  feet  4  inches,  and  breadth 
8  feet  3  inches  ?  '  Ans.  5/.  7s.  I  Urf. 

Exam.  2.  What  cost  the  paving  a  court,  at  3s.  2d.  per 
yard  ;  the  length  being  2  7  feet  10  inches,  and  the  breadth 
14  feet  9  inches  ?  .  Ans.  11.  4s.  b^d. 

Exam.  3.  What  will  be  the  expense  of  paving  a  rectan- 
gular court-yard,  whose  length  is  63  feet,  and  breadth  45 
feet  ;  in  which  there  is  laid  a  foot-path  of  5  feet  3  inches 
broad,  running  the  whole  length,  with  broad  stones,  at  3s. 
a  yard  ;  the  rest  being  paved  with  pebbles  at  2s.  6d.  a  yard  ? 

Ans.  40/.  5s.  iOic/. 


XI.  PLUMBERS'  WORK. 

Plumbers'   work  is  rated  at  so  much  a  pouifd',  or  else  by 

the  hundred  weight  of  112  pounds- 
Sheet  lead,  used  in  roofing,  guttering,    5lc.    is  from  6  to 

10  lb.  to  the  square  foot.     And  a  pipe   of  an  inch  bore  is 

aommonly  13  or  14  lb.  to  the  yard  in  length. 

EXAMPLES. 

•  Exam.  1.     How  much  weighs  the  lead  which  is  39  feet 
Vol..   I.  69  6  inches 


466  TIMBER  MEASURING. 

6  inches  long,  and  3  feet  3   inches  broad,  at  Silfe.  to  the 
square  foot  ?  Ans.  109lyyb, 

Exam.  2.  What  cost  the  covering  and  guttering  a  roof 
with  lead,  at  18s.  the  cwt.  ;  the  length  of  the  roof  being  43 
feet,  and  breadth  or  girt  over  it  32  feet  ;  the  guttering  57 
feet  long,  and  2  feet  wide  ;  the  former  9*831  lb.  and  the  lat- 
ter 7-37^  ib.  to  the  square  foot  ?  Ans.  116/.  9s,  Urf. 


XII.  TIMBER  MEASURING. 

PROBLEM  I, 
To  find  the  Area^  or  Superficial  Content^  of  a  Board  or  Plank. 

Multiply  the  length  by  the  mean  breadth. 

J\'ote.  When  the  board  is  tapering,  add  the  breadths  at 
the  two  ends  together,  and  take  half  the  sum  for  the  mean 
breadth.     Or  else  take  the  mean  breadth  in  the  middle. 

By  the  Sliding  Rule. 

Set  12  on  b  to  the  breadth  in  inches  on  a  :  then  against 
the  length  in  feet  on  b,  is  the  content  on  a,  in  feet  and 
fractional  parts. 

EXAMPLES. 

Ej^m.  1.  What  is  the  value  of  a  plank,  at  l^d.  per  foot, 
whose  length  is  12  feet  6  inches,  and  mean  breadth  11  inches  ? 

Ans.  Is.  5d. 
Exam.  2.     Required  the  content  of  a  board,  whose  length 
is  11  feet  2  inches,  and  breadth  1  foot  10  inches  ? 

Ans   20  feet  b  inches  8''. 
Exam.  S.     What  is  the  value  of  a  plank,  which  is  12  feet 
9  inches  long,  and  1  foot  3  inches  broad,  at  2irf.  a  foot  ? 

Ans.  3s  3|d. 
Exam.  4.  Required  the  value  of  5  oaken  planks  at  3d. 
per  foot,  each  of  them  being  17|  feet  long  ;  and  their  several 
breadths  as  follows,  namely,  two  of  1.3^  inches  in  the  middle, 
one  of  14i  inches  in  the  middle,  and  the  two  remaining 
ones,  each  18  inches  at  the  broader  end,  and  1  4  at  the  nar- 
rower ?  Ans.  1/.  5s.  9id. 

PROBLEM 


TIMBER  MEASURING.  467 


PROBLEM  U. 


To  find  the  Solid  Content  of  Squared  or  Four- sided  Timber, 

Multiply  the  mean  breadth  by  the  mean  thickness,  and  the 
product  again  by  the  length,  for  the  content  nearly. 

By  the  Sliding  Rule. 

CD  DC 

As  length  :   12  or  10  :  :  quarter  girt :  solidity. 

That  is,  as  the  length  in  feet  on  c,  is  to  12  on  ©,  when 
the  quarter  girt  is  in  inches,  or  to  10  on  d,  when  it  is  in 
tenths  of  feet  ;  so  is  the  quarter  girt  on  d,  to  the  content 
on  c. 

JVote  1.  If  the  tree  taper  regularly  from  the  one  end  to 
the  other  ;  either  take  the  mean  breadth  and  thickness  in  the 
middle,  or  take  the  dimensions  at  the  two  ends,  and  half 
their  sum  will  be  the  mean  dimensions  ;  which  multiplied  as 
above,  will  give  the  content  nearly. 

2.  If  the  piece  do  not  taper  regularly,  but  be  unequally 
thick  in  some  parts  and  small  in  others  ;  take  several  diffe- 
rent dimensions,  add  them  all  together,  and  divide  their  sum 
by  the  number  of  them,  for  the  mean  dimensions. 

EXAMPLES. 

Exam.  1.  The  length  of  a  piece  of  timber  is  18  feet  6  in- 
ches, the  breadths  at  the  greater  and  less  end  1  foot  6  inches 
and  1  foot  3  inches,  and  the  thickness  at  the  greater  and  less 
end  1  foot  3  inches  and  1  foot ;  required  the  solid  content  ? 

Ans.  28  feet  7  inches. 

Exam.  2.  What  is  the  content  of  the  piece  of  timber, 
whose  length  is  24|  feet,  and  the  mean  breadth  and  thickness 
each  1-04  feet  ?  .  Ans.  26^  feet. 

Exam.  3.  Required  the  content  of  a  piece  of  timber, 
whose  length  is  20  38  feet,  and  its  ends  unequal  squares,  the 
sides  of  the  greater  being  191  inches,  and  the  side  of  the  less 
91  inches  ?  Ans.  28-7562  feet. 

Exam. 


468  TIMBER  MEASURING. 

Exam.  4.  Required  the  content  of  the  piece  of  timber, 
whose  length  is  27 -36  feet  ;  at  the  greater  end  the  breadth  is 
1-78,  and  thickness  1  23  ;  and  at  the  less  end  the  breadth  is 
1-04,  a^d  thickness  0'91  feet  ?  Ans.  41*278  feet. 

PROBLEM  III. 

To  find  the  Solidity  of  Round  or  Unsquared  Timber. 

Multiply  the  square  of  the  quarter  girt,  or  of  ^  of  the 
mean  circumference,  by  the  length,  for  the  content. 

By  the  Sliding  Rule. 

As  the  length  upon  c  :  12  or  10  upon  d  :  : 
quarter  girt,  in  12ths  or  lOths,  on  d  :  content  on  c* 

JVoi^e  1.  When  the  tree  is  tapering,  take  the  mean  dimen- 
sions as  in  the  former  problems,  either  by  girting  it  in  the 
middle,  for  the  mean  girt,  or  at  the  two  ends,  and  take  half 
the  sum  of  the  two  ;  or  by  girting  it  in  several  places,  then 
adding  all  the  girts  together,  and  dividing  the  sum  by  the  num- 
ber of  them,  for  the  mean  girt.  But  v.'hen  the  tree  is  very 
irregular,  divide  it  into  seyeral"  lengths,  and  tind  the  content 
of  each  part  separately. 

2.  Tiiis  rule,  which  is  commonly  used,  gives  the  answer 
about  i  less  than  the  true  quantity  in  the  tree,  or  nearly  what 
the  quantity  would  be,  after  th(i  tree  is  hewed  square  in  the 
usual  way  :  so  that  it  seems  intended  to  make  an  allowance 
for  the  squaring  of  the  tree. 

EXAMPLES. 

Exam.  1.  A  piece  of  round  timber  being  9  feet  0  inches 
long,  and  its  mean  quarter  girt  42  inches  ;  what  is  the  con- 
tent ?  Ars.  11  6J  feet. 

Exam.  2.  The  length  of  a  tree  is  24  feet,  its  girt  at  the 
thicker  end  14  feet,  and  at  the  smaller  end  2  feet  ;  required 
the  content  ?  Ans.  96  feet. 

Exam.  3.  What  is  the  content  of  a  tree,  whose  mean  girt 
is  3-15  feet,  and  length  14ieet  6  inches  ? 

Ans.  8-9922  feet. 

Exam.  4.  Required  the' content  of  a  tree,  whose  length 
is  17i  feet,  which  girts  in  tive  different  places  as  follows, 
namely,  in  the  tirst  place  9-43  feet,  in  the  second  7-92,  in  the 
(bird  615.  in  the  fourth  4-74,  and  in  the  fifth  3- 16  ? 

Ans.  42-619525. 
CONIC 


[469] 


eONIC  SECTIONS. 


DEFINITIONS. 


1.  Conic  Sections  are  the  figures  made  by  a  plane  cuttiag 
a  cone. 

2.  According  to  the  diflferent  positions  of  the  cuttins:  plane, 
there  arise  five  different  iigures  or  sections,  namely,  a  triiangle, 
a  circle,  an  ellipsis,  an  hyperbola,  and  a  parabola  :  the  three 
last  of  which  only  are  peculiarly  called  Conic  Sections. 

vT 

3.  If  the  cutting  plane  pass  through 
the  vertex  of  the  cone,  and  any  part  of 
the  base,  the  section  will  evidently  be  a 
triangle  ;  as  vab. 


4.  If  the  plane  cut  the  cone  parallel  t« 
the  base,  or  make  no  angle  with  it,  the 
section  will  be  a  circle  ;  as  abd. 


5.  The  section  dab  is  an  ellipse  when 
the  cone  is  cut  obliquely  through  both 
sides,  or  when  the  plane  is  inclined  to 
the  base  in  a  less  angle  than  the  side  of 
the  cone  is. 


6.  The  section  is  a  parabola,  when 
the  cone  is  cut  by  a  plane  parallel  to 
the  side,  or  when  the  cutting  plane  and 
the  side  of  the  cone  make  equal  angles 
with  the  base. 


7.  The 


470 


GONIC  SECT[0NJS. 


7.  The  section  is  an  hyperbola,  when 
the  cutting  plane  makes  a  greater  angle 
with  the  base  than  the  side  of  the  cone 
makes. 


8.  And  if  all  the  sides  of  the  cone 
be  continued  through  the  rertex,  form- 
ing an  opposite  equal  cone,  and  the 
plane  be  also  continued  to  cut  the  op- 
posite cone,  this  latter  section  will  be 
the  opposite  hj'perbola  to  the  former  ; 
as  dee. 


And  further,  if  there  be  four  cones 
GMN,  COP,  CMP.  cNo,  having  all  the  same 
vertex  c,  and  all  their  axes  in  the  same 
plane,  and  their  sides  touching  or  coin- 
ciding in  the  common  intersecting  lines 
Mco,  NOP  ;  then  if  these  four  cones  be 
all  cut  by  one  plane,  parallel  to  the  com- 
mon plane  of  their  axes,  there  will  be 
formed  the  four  hyperbolas  rgq,  sft, 
KVL,  HWi,  of  which  each  two  opposites 
are  equal,  and  the  other  two  are  con- 
jugates to  them  ;  as  here  in  the  annexed 
figure,  and  the  same  as  represented  in  the 
two  following  pages. 

9.  The  Vertices  of  any  section,  are  the  points  where  the 
cutting  plane  meets  the  opposite  sides  of  the  cone,  or  the  sides 
of  the  vertical  triangular  section  ;  as  a  and  b. 

Hence  the  ellipse  and  the  opposite  hyperbolas,  have  each 
two  vertices  ;  but  the  parabola  only  one  ;  unless  we  consider 
the  other  as  at  an  infinite  distance.- 

10.  The  Axis,  or  Transverse  Diameter,  of  a  conic  section, 
is  the  line  or  distance  ab  between  the  vertices. 

F   Hence  the  axis  of  a  parabola  is  infinite  in  length,  Ab  being 
only  a  part  of  it. 


Ellipse. 


DEFINITIONS. 


471 


Ellipse. 


Hyperbolas. 


Parabola. 


11.  The  Centre  c  is  the  middle  of  the  axis. 

Hence  the  centre  of  a  parabola  is  infinitely  distant  from  the 
rertex.  And  of  an  ellipse,  the  axis  and  centre  lie  within  the 
curve  ;  but  of  an  hyperbola,  without. 

12  A  Diameter  is  any  right  line,  as  ab  or  de,  drawn 
through  the  centre,  and  terminated  on  each  side  by  the  curve; 
and  the  extremities  of  the  diameter,  or  its  intersections  with 
the  curve,  are  its  vertices. 

Hence  all  the  diameters  of  a  parabola  are  parallel  to  the 
axis,  and  infinite  in  length.  And  hence  also  every  diameter 
of  the  ellipse  and  hyperbola  have  two  vertices  ;  but  of  the 
paraboli,  only  one  ;  unless  we  consider  the  other  as  at  an  in- 
finite distance. 

13.  The  Conjugate  to  ariy  diameter,  is  the  line  drawn 
through  the  centre,  and  parallel  to  the  tangent  of  the  curve 
at  the  vertex  of  the  diameter.  So,  fg,  parallel  to  the  tangent 
at  D,  is  the  conjugate  to  de  ;  and  hi,  parallel  to  the  tangent 
At  A,  is  the  conjugate  to  ab. 

Hence  the  conjugate  hi,  of  the  axis  ab,  is  perpendicular 
to  it. 

14.  An  Ordinate  to  any  diameter,  is  a  line  parallel  to  its 
conjugate,  or  to  the  tangent  at  its  vertex,  and  terminated  by  the 
diameter  and  curve.  So  dk,  el,  are  ordinates  to  the  axis  ab  ; 
and  MN,  NO,  ordinates  to  the  diameter  de. 

Hence  the  ordinates  of  the  axis  are  perpendicular  to  it. 

15.  An  Absciss  is  a  part  of  any  diameter  contained  be- 
tween its  vertex  and  an  ordinate  to  it  ;  as  ak  or  bk,  or  dn 
dr  EN. 

Hence,  in  the  ellipse  and  hyperbola,  every  ordinate  has  two 
determinate  abscisses  ;  but  in  the  parabola,  only  one  ;  the  other 
vertex  of  the  diameter  being  infinitely  distant. 

16.  The  Parameter  of  any  diameter,  is  a  third  proportion- 
al to  that  diameter  and  its  conjugate. 

17.  The 


472 


CONIG  SECTION^: 


17.  The  Focus  is  the  point  in  the  axis  where  the  ordinate 
is  equal  to  half  the  parameter.  As  k  and  l,  where  dk  or  el 
is  equal  to  the  semi- parameter.  'J  he  name  focus  heing  given 
to  this  point  from  the  pecuHar  properly  of  it  mentioned  in  the 
corol  to  theor.  9  in  the  Ellipse  and  Hyperbola  following,  and 
to  theor.  6  in  the  Parabola. 

Hence,  the  ellipse  and  hyperbola  have  each  two  foci  ;  but 
the  parabola  only  one.  «. 


18.  If  DAK,  FBG,  be  two  opposite  hyperbolas,  having  ae 
for  their  first  or  tran*vei^e  axis,  and  ab  for  their  second  or 
conjugate  axis.  And  if  dae,  fbg,  be  two  other  opposite  hy- 
perbolas having  the  same  axes,  but  in  the'contrary  order, 
namely,  ab  their  first  axis,  and  ab  their  second  ;  then  these  two 
latter  curves  dae,  fbg,  are  called  the  conjugate  hyperbolas  to 
the  two  former  dae,  fbg,  and  each  pair  of  opposite  curves 
mutually  conjugate  to  the  other  ;  being  all  cut  by  one  plane, 
from  four  conjugate  cones,  as  in  page  470,  def.  8. 

19.  And  if  tangents  be  drawn  to  the  four  vertices  of  the 
curves,  or  extremities  of  the  axes,  forming  the  inscribed 
rectangle  hikl  ;  the  diagonals  hck,  icl,  of  this  rectangle, 
are  called  the  asymptotes  of  the  curves.  And  if  these  asymp- 
totes intersect  at  right  angles,  or  the  inscribed  rectangle  be 
a  square,  or  the  two  axes  ab  and  ab  be  equal,  then  the  hyper- 
bolas are  said  to  be  right-angled,  or  equilateral: 

SCHOLIUM. 

The  rectangle  inscribed  between  the  four  conjugate  hy- 
perbolas, is  similar  to  a  rectangle  circumscribed  about  an 
ellipse,  by  drawing  tangents,  in  like  manner,  to  the  four  ex- 
tremities of  the  two  axes  ;  and  the  asymptotes  or  diagonals 
in  the  hyperbola,  are  analogous  to  those  in  the  ellipse,  cut- 
ting this  curve  in  similar  points,  and  making  that  pair  of 
conjugate  diameters  which  are  equal  to  each  other  Also, 
the  whole  figure  formed  by  the  four  hyperbolas,  is,  as  it 
were,  an  ellipse  turned  inside  out,  cut  open  at  the  extre- 
mities D,  E,  F,  G,  of  the  said  equal  conjugate  diameters,  and 
those  four  points  drawn  out  to  an  infinite  distance  ;  the  cur- 
vature being  turned  the  contrary  way,  hnt  the  axes,  and  the 
rectangle  passing  through  their  extremities,  continuing  fixed. 

©F 


OF  THE  ELLIPSE.  473 


OF  THE  ELLIPSE, 


THEOREM  I, 


The  Squares  of  the  Ordinates  of  the  Axis  are  to  each  other 
as  the  Rectangles  of  their  Abscisses. 

Let  avb  be  a  plane  passing  through  V 

the   axis  of  the  cone  ;    agih   another  /V 

flection  of  the   cone   perpendicular  to  /   \ 

the  plane   of  the  former  ;  ab  the  axis  /^^""y^ 

of  this  elliptic  section  ;  and  fg,  hi,  or-  ^^.JI^JJj 

dioates  perpendicular  to  it.     Then  it  wS^^l^^AjT 

will  be,  as  fg2  :  hi*  :  :  af  .  fb  :  ah  .  hb.  ■^^^^^^^^^'^^^^ 

For,  through  the   ordinates  fg,  hi, 
draw  the  circular  sections   kgl,    min, 

parallel  to  the  base  of  the  cone,  having  kl,  mn,  for  their  di- 
ameters, to  which  fg,  hi,  are  ordinates,  as  well  as  to  the  axii& 
of  the  ellipse. 

]N^ow,  by  the  similar  triangles  afl,  ahn,  and  bfe,  shjS^ 

it  is  AF  :  ah  :  :  pl  :  hn,    • 
and  FB  :  HE  :  :  kf  :  mh  ; 

hence,  taking  the  rectangles  of  the  corresponding  termji, 
it  is,  the  rect.  af  .  fb  :  ah  .  hb  :  :  kf  .  fl  :  mh  .  hn. 

3ut,  by  the  circle,  kf  .  fl  =  fg^,  and  mh     hn  =  hi*  ; 
Therefore  the  rect.  af  .  fb  :  ah  .  hb  ;  :  f«*  :  hi*.  ^.  £.  ». 


THEOREM  11. 


As  the  Square  of  the  Transverse  Axis  ; 
Is  to  the  Square  of  the  Conjugate     :  ; 
ISo  is  the  Rectangle  of  the  Abs.'.issejS     :» 
To  the  Square  ef  their  Ordinate. 

?e».  }.  61  .  Tkat 


4f4 


CONIC  SECTIONS. 


That  is,  ab2  :  ab^  or 
Ac2  :  ac^  :  :  AD  .  db  :  de^. 


For,  by  theor.  1 ,  ac  .  cb  :  ad  .  db  :  :  ca^  :  de^  ; 

But,  if  c  be  the  centre,  then  ac  .  cb  =»  ac^  ,  and  ca  is  the 

semi-conjugate. 
Therefore  ac^  :  ad  .  db  :  :  ac^  :  de^  ; 

or,  by  permutation,  ac^  :  ac^  :  :  ad  .  db  :  de^  ; 
or,  by  doubling,        ab^  :  ab«  :  :  ad  .  db  :  de^.  q.  e.  ». 

ab» 
Corol.  Or,  by  div.  ab  :  • —  :  :  ad  .  db  or  ca^  —  cd^  :  de*  , 
ab 
that  is,  AB  :  2?  :  :  AD  .  db  or  ca^  —  cd^  :  de^  ; 
aba 
where  p  is  the  parameter  -^-^j  by  the  definition  of  it. 

AB 

That  is.  As  the  transverse. 
Is  to  its  parameter, 
So  is  the  rectangle  of  the  abscisses, 
To  the  square  of  their  ordinate. 

THEOREM  m. 


As  the  Square  of  the  Conjugate  Axis  : 

Is  to  the  Square  dof  the  Transverse  Axis  :  : 

So  is  the  Rectangle  of  the  Abscisses  of  the  Conjugate,  or 
the  Difference  of  the  Squares  of  the  Semi-conjugate  and 
Distance  of  the  Centre  from  any  Ordinate  of  that  Axis  : 
To  the  Square  of  their  Ordinate. 


ca3  :  cp" 


That  is, 
:  :  ad  .  db  or  ca^—cd*  :  dE^. 


For,  draw  the  ordinate  ed  to  the  transverse  ab. 
Then,  by  theor.  2,  cas  :  ca^  : :  de^  :  ad  .  db  ©r  ca^  — cd^, 
or  -     -     -     -     ca?  ;  ca3  : :  cd^  :  cA^—dE^. 

But        -     -     -     -    ca2  :  ca^  :  :  ca^  :  ca^, 
theref.  by  subtr.         ca^  :  ca^  :  :  ca*  —cd^  or  ad  .  db  :  de^. 

d.  E.  d. 
Corol. 


OF  THE  ELLIPSE.  475 

CoroL  1.  If  two  circles  be  described  on  the  two  axes  as 
diameters,  the  one  inscribed  within  the  ellipse,  and  the  other 
circumscribed  about  il ;  then  an  ordinate  in  the  circle  will 
be  to  the  corresponding  ordinate  in  the  ellipse,  as  the  axia  of 
this  ordinate,  is  to  the  other  axis. 

That  is,  CA  :  ca  :  :  dg  :  de, 
and  CA  :  CA  :  :  dg  :  dE. 
For,  by  the  nature  of  the  circle,  ad  .  db  =  dg^  ;  theref. 
by  the  nature  of  the  ellipse,  ca^  :  ca3  : :  ad  .  db  or  dg^  :  de^, 
or  CA  :  ca  :  :  dg  :  de. 
In  like  manner  -       ,  ca  :  ca  :  :  dg  :  ds. 

Also,  by  equality,      -         dg  :  de  or  cd  :  :  dE  or  dc  :  dg. 
Therefore  cgG  is  a  continued  strai|ht  line. 

Corol.  2.  Hence  also,  A  the  ellipse  and  circle  are  made  up 
of  the  same  number  of  corresponding  ordinates,  which  are 
all  in  the  same  proportion  of  the  two  axes,  it  follows  that 
the  areas  of  the  whole  circle  and  ellipse,  as  also  of  any  like 
parts  of  them,  are  in  the  same  proportion  of  the  two  axes, 
or  as  the  square  of  the  diameter  to  the  rectangle  of  the  two 
axes  ;  that  is,  the  areas  of  thfi  two  circles,  and  of  the  ellipse, 
are  as  the  square  of  each  axis  and  the  rectangle  of  the  two"^ 
and  therefore  the  ellipse  is  a  mean  proportional  between  the 
two  circles. 


THEOREM  rV. 

The  Square  of  the  Distance  of  the  Focus  from  the  Centre, 
is  equal  to  the  Difference  of  the  Squares  of  the  Semi- 
axes  ; 

Or,  the  Square  of  the  Distance  between  the  Foci,  is  equal  to 
the  Difference  of  the  Squares  of  the  two  Axes. 


That  is,  cf2  =  ca^  —  ca^, 
or  Ff3  =  ab2  —  ab2. 


b 

For,  to  the  focus  f  draw  the  ordinate  fe  ;  which,  by  the 
definition,  will  be  the  semi-parameter.     Then  by  the  nature 
of  the  curre       -         -         ca^  :  ca^  : :  ca^  —  cf^  :  fb^  ; 
and  by  the  def.  of  the  para,  ca^  :  ca^  ::        ca^       :  fe^  ; 
therefore  -         -         ca^  =ca2  —  cf^  ; 

and  b)  addit.  and  subtr.         cf^  =ca2  —  ca*  ; 
or,  by  doubling,  -         fP=ab3 —abs  ;  <i.  e.  d. 

Corol. 


476  CO^tC  SECTIONS. 

Corot.  1.  The  two  semi-axes,  and  the  focal  distance  from 
the  centre,  are  the  sides  of  a  right  angled  triangle  era  ;  and 
the  distance  Fa  from  the  focus  to  the  extremity  of  the  conju- 
gate axis,  is  =  AC  the  semi-transverse. 

Corol.  2.  The  conjugate  semi-axis  ca  is  a  mean  propor- 
tional between  af,  fb,  or  between  Af,  fe,  the  distances  of  ei- 
ther focus  from  the  two  vertices. 

For  ca3  s=  ca^  -cf^  =  (ca  -|-  cf)  .  (ca  -  cf)  s=  af  .  fb'V 
THEOREM  V. 

The  sum  of  two  Lines  drawn  from  the  two  Foci  to  meet  at 
any  Point  in  the  Curve,  is  equal  to  the  Transverse  Axis. 


That  is, 
f  E  -H  fe  =  ^ 


For,  draw  ag  parallel  and  equal  to  ca  the  semi-conjugate  ; 
»nd  join  cg  meeting  the  ordinate  de  in  h  ;  also  take  ci  a  4th 
proportional  to  ca,  cf,  cd. 

Then,  by  theor.  2,  ca2  :  ag^  :  :  ca^  — cd^^  :  de^  ; 

and,  by  sim.  tri.        ca^  :  ag2  :  :  ca^  — cd^  :  ag^  —  dh^  * 

consequently  de^  =  ag^  — dh^  =ca2  —  dh^. 

Also  FD  =  cf  (/3  CD,  and  fd^  =  cf^  —  2cf  .  cd  4"  cd'  ; 
and,  by  right-angled  triangles,      fe^  =  fd^  -f-  de^  ; 
therefore  fe^  =  cf^  +  ca^  —  2cf  .  cd  -j-  cd^  —  dh^. 

But  by  theor.  4,  cf2  +  ca2  =  ca^, 

and  by  supposition,      2cf  .  cd=  2ca  .  ci  ; 
theref.  fe^  =  ca^  —  2ca  .  ci  -f-  cd^  —  dh^ 

Again,  by  supp.    ca^  :  cd^  :  :  cf^  or  ca^  — ag2  ci^  ; 

and,  by  sim.  tri.  ca^  :  cd^  :  :  ca2  — ag2  :  cd*  •—  dh^  : 

therefore        -       ci2  =  cd2  —  dh*  ; 

consequently         fe^  =  ca2  — 2ca  .  ci  -f-  ci®. 

And  the  root  or  side  of  this  square  is  fe  =  ca  »—  ci  =  ai. 

In  the  same  manner  it  is  found  that  fE  =  ca  -f-  ci  =  bi. 
Conseq.  by  addit.  fe  -f-  fE  =  ai  -|-  bi  :=  ab.  q.  e.  p*. 


OP  THE  ELLIPSE. 


477 


€orol.  1.  Hence  ci  or  ca 


FE  is  a  4th  proportional  to  ca, 

OF,  CD. 

Corol.  2  And  fc  — fe  =  2oi  ;  that  is,  the  difference  be- 
tween two  lines  drawn  from  the  foci,  to  any  point  in  the  curve, 
is  double  the  4th  proportional  to  ca,  cf,  cd. 

Corol.  3.  Hence  is  derived  the  common  method  of  de- 
icribing  this  curve  mechanically  by  points,  or  with  a  thread 
thus  : 

In  the  transverse  take  the  foci  f,  f, 
and  any  point  i.  Then  with  the  radii 
Ai,  Bi,  and  centres  f,  f,  describe  arcs 
intersectinj;  in  e^,  which  will  be  a 
point  in  the  curve.  In  like  manner, 
assuming  other  points  i,  as  many 
other  points  will  be  found  in  the 
curve.  Then  with  a  steady  hand 
the  curve  line  may  be  drawn  through  all  the  points  of  inter- 
section E. 

Or,  take  a  thread  of  the  length  ab  of  the  transverse  axis, 
and  fix  its  two  ends  in  the  foci  f,  f,  by  two  pins.  Then  carry 
a  pen  or  pencil  round  by  the  thread,  keeping  it  always  stretch- 
ed, and  its  point  will  trace  out  the  curve  line. 

THEOREM  VL 

Tf  from  any  Point  i  in  the  Axis  produced,  a  Line  il  be 
drawn  touching  the  curve  in  one  point  l  ;  and  the  Ordi- 
nate LM  be  drawn  ;  and  if  c  be  the  Centre  or  Middle  of 
ab  :  Then  shall  cm  be  to  ei  as  the  Square  of  abi  to  the 
Square  of  ai. 


That  is, 
CM  :  ci  :  :  am^ 


For,  from  the  point  i  draw  any  other  line  ieh  to  cut  the 
curve  in  two  points  e  and  h  ;  from  which  let  fall  the  perpen- 
diculars ED  and  HG  ;  and  bisect  dg  in  k. 

Then,  by  theo    1,  ad  .  db  :  ag  .  gb  :  de^  :  gh^, 
and  by  sim.  triangles,    id^  :  ig^  :  :  de^  :  gh^  ; 
theref.  by  equality,  ad  ,  db  :  ag  .  gb  :  :  id*  :  ig^. 

But  DB  =  CB  4-  CD  =  AC  4-  CD  =  AG  -f"  DC  —  CG  =  2cK  -f  AG, 

and  GB  =  CB  —  CG  =  ac—  cg  =  ad  4"  dc  —  cg  =  2ck  4-  ad  ; 

theref.. AD .  2ck  4"  ad  .  ag  :  ag  .  2ck  4-  ad  ,  ag    :  :  id^  :  ig^, 

and,bydiv.  DG  .  2ck  :  lo^-^ips  ^^y  p^  .  gjg  ;  .   ad  -  2ck  4* 

ad  .  AG  :  i»3,  or 


476 


CONIC  SECTIONS. 


or     -     2eK :  2iK  :  :  AD  .  2ck  4*  ad  .  aq  :  id^  , 

or     AD  .  2cK  :  ad  .  2ik  :  :  ad  :  2ck  -f  ad  .  ag  :  id^  ; 

theref.  by  div.  ck  :  ik  :  :  ad   ag  :  id^— ad  :  ik 


and,  by  comp    ck  :  ic  :  :  ad   ag  :  id^-^ad  .  id  -f-  ia, 
or         -         CK  :  CI  :  :  AD  .  AG  :  ai^. 

But,  when  the  line  m,  by  revolving  about  the  point  i,  comes 
into  the  position  of  the  tangent  il,  then  the  points  e  and  h 
meet  in  the  point  l,  and  the  points  d,  k,  g,  coincide  with  the 
point  M ;  and  then  the  last  proportion  becomes  cm  :  ci : :  am^  :  ai^. 

a.  E.  D* 


THEOREM  VII. 


If  a  Tangent  and  Ordinate  1?e  drawn  from  any  Point  in  the 
Curve,  meeting  the  Transverse  Axis  ;  the  Semi-transverse 
will  be  a  Mean  proportional  between  the  Distances  of  the 
said  Two  Intersections  from  the  Centre. 


Th^itis, 
CA  is  a  mean  proportional  be- 
tween CD  and  CT  ; 
or  CD,  CA,  CT,  are  continued 
proportionals. 


For,^  by  theor.  6,  cd  :  ct  :  :  ad^  :  at^. 
that  is,  CD  :  cT  :  :  (ca  — cd)2  :  (cT— Oa)^, 

or  -         CD  :  CT  :  :  cd^  -{-ca^  :  ca^  -{-  ct^  ,  ' 

and         -         CD  :  DT  :  :  cD^-f-CA^  :  ct^  —  cd^, 
or  -         CD  :  DT  :  :  cd^  -f  ca^  :  (ct  -{-  gd)  .  dt, 

or  -         cd2  :  CD  .  dt  :  :  cd^  +  ca^  :  cd  .  dx  +  ct  .  dt, 

hence     -         cd^  :  ca^  :  ;  cd  .  dt  :  ct  .  dt, 
and  -         cd2  :  ca^  :  : cd  f  ct. 

therefore  (th.  78,  Geom.)  c©  :  ca  :  :  ca  :  ct.  ft.  e.  d. 

Corol.  Since  ct  is  always  a  third  proportional  to  cd,  ca  ; 
if  the  points  d,  a,  remain  constant,  then  will  the  point  t  be 
constant  also  ;  and  therefore  all  the  tangents  will  meet  in 
this  point  t,  which  are  drawn  from  the  point  e,  of  every 
ellipse  described  on  the  same  axis  ab,  where  they  are  cut. by 
the  common  ordinate  dee  drawn  from  the  point  p. 

THEOREM 


OF  THE  ELLIPSE. 


47S 


THEOREM  Vin. 

If  there  be  any  Tangent  meeting  Four  Perpendiculars  to  the 
Axis  drawn  from  these  four  Points,  namely,  the  Centre,  the 
two  Extremites  of  the  Axis,  and  the  Point  of  Contact  ;  those 
Four  Perpendiculars  will  be  Proportionals. 

11 


That  is, 
AG  :  DE  :  :  CH  :  bi. 


For,  by  theor.  7,  tc  :  ac  :  :  ac  :  dc, 
theref  by  div.  ta  :  ad  :  :  Tq  :  ac  or  cb, 

and  by  comp.  ta  ;  td  :  :  tc  :  tb, 

and  by  sim.  tri.         ag  :  de  :  :  ch  :  bi.  ,  q;  e.  ». 

Carol.  Hence  ta,  td,  tc,  tb  >  ,  *•       i 

and      TG,  TE,  TH,  T,  |  "«  ^^'^  ProportioDaU. 
For  these  are  as  ag,  ©e,  ch,  bi,  by  similar  triangles. 

THEOREM  IX. 

If  there  be  any  Tangent,  and  two  Lines  drawn  from  the  Foci 
to  the  Point  of  Contact  ;  these  two  Lines  will  make  equal 
Angles  with  the  Tangent* 

7e 


That  is, 
the  ^  FET  =  ZfEe. 


For,  draw  the  ordinate  de,  and  fe  parallel  to  fe 
By  cor.  1,  theor.  5,     ca  :  cd  :  :  cf  :  ca  ^ —  fe, 
CA  :  CD  :  :  CT  :  CA  ; 
CT  :  cp  :  :  CA  ;  ca  — ^  fe  ; 
TF  :  Tf  :  :  FE  :  2ca  —  fe  or Te  by  th.  5. 
TF  :  Tf  :  :  FE  :  fe  ; 

fE  =  fe,  and  conseq.  Ze  =  ^fse. 

FE  is  parallel  to  fe,  the  ^e  =  Z^et  ; 
Zfet  =  Zft^G.  Q.  E.  D. 

CQrol. 


and  by  theor.  7, 

therefore 

and  by  add.  and  sub. 

But  by  sim.  tri. 

therefore 

But,  because 

therefore  the 


480  CONIO  SECTIONS. 

Corol.  As  opticians  find  that  the  angle  of  incidence  is  equal 
to  the  angle  of  reflection,  it  appears  from  this  theorem,  that 
rays  of  light  issuing  from  the  one  focus,  and  meeting  th^  curve 
in  every  point,  will  be  reflected  into  lines  drawn  from  those 
points  to  the  other  focus.  So  the  ray  {e  is  reflected  into  fe. 
And  this  is  the  reason  why  the  points  f,  f,  are  called  the/oa, 
or  burning  points. 

THEOREM  X. 

All  the '  Parallelograms  circumscribed  about  an  Ellipse  arc 
equal  to  one  another,  and  each  equal  to  the  Rectangle  of 
the  two  Axes. 


That  is, 
the  parallelogram  pqrs  = 
the  rectangle  ab  .  ab. 


S 
Let  EG,  eg,  be  two  conjugate  diameters  parallel  to  the  sides 
of  the  parallelogram,  and  dividing  it  into  four  less  and  equal 
parallelograms.  Also,  draw  the  ordinates  de,  de,  and  ck  per- 
pendicular to  PQ  ;  and  let  the  axis  ca  produced  meet  the  sides 
of  the  parallelogram,  produced  if  necessary,  in  t  and  t. 
Then  by  theor  7, 
and 

theref  by  equality, 
but,  by  sim.  triangles, 
theref  by  equality, 
and  the  rectangle 
Again,  by  theor.  7, 
or,  by  division, 
and  by  composition, 
conseq.  the  rectangle 

But,  by  theor.  2,  ca^  :  ca^  :  :  (ad  .  db  or)  cd^  :  de^, 

therefore 


*Corol  Because  c^^  ■«  AD  .  db  —  ca*  -cd*, 
therefore     ca^  «»  gd*  4-  cd^. 
In  like  maimer,      ca^  <=  jde^  -f  de®. 

Id 


CT 

CA  : 

CA  :  CD, 

ct 

CA  : 

CA  :  cd  ; 

CT 

ct  : 

cd  :  CD  ; 

CT 

Ct  : 

TD  :  cd, 

TD 

cd  : 

CD  :  CD, 

TD 

DC  is 

=  the  square  cd^. 

CD 

CA  : 

CA  :  CT, 

CD 

CA  :  : 

DA  :  AT, 

CD 

DB  : 

AD  :  DT  ; 

CD 

DT  = 

-  Cd2  =  AD  .  DB*. 

CA2 

:  ca2 

:  :  (ad  .  DB  or)  cd^ 

ca 

ca  :  : 

cd  :  DE  ; 

OF  THE  ELLIPSE. 


48t 


Iq  like  manner, 

«r 

But,  by  theor.  7, 

theref.  by  equality, 

But,  by  sim.  tri. 

theref.  by  equality, 

and  the  rectangle  ck  .  ce  =  ca  .  ca. 

But  the  rect.         ck  .  ce  =  the  parallelogram  cepc, 

theref  the  rect.    ca  .  ca  =  the  parallelogram  cEpe, 

eonseq.  the  rect.  ab  .  ab  =  the  parallelogram  pqrs. 


CA 

:  ca  : 

:  CD 

:  de, 

ca 

de  : 

:  CA 

CD. 

CT 

:  CA  : 

:  ca 

CD  , 

de. 

CT 

CA  : 

:  ca 

CT 

CK  : 

:  ce 

.  de 

CK 

CA  : 

:  ca 

ce, 

THEOREM  XL 

The  Sum  of  the  Squares  of  every  Pair  of  Conjugate  Diame- 
ters, is  equal  to  the  same  constant  (Quantity,  namely,  the 
Sum  of  the  Squares  of  the  two  Axes. 


That  is, 
AB^+ab^  =  eg2  -f-  eg2  ; 
where  eg,  eg,  are  any  pair  of  con- 
jugate diameters. 


For,  draw  the  ordinates  ed,  ed. 
Then,  by  cor  to  theor.   10,  ca^  =  cd^  -f*  cd*» 
and         .         -         -  ca2  =  de-  -j-  de^  ; 

therefore  the  sum       ca^  -\-  ca*  =  cd*  +  de^  -f.  cd^ 
But,  by  right-angled  As,       ce*  =  cd*  -f-  de*, 
and         .         -         -  ce*  =  cd*  -f  de*  ; 

therefore  the  sum       ce*  +  ce*  =  cd*  +  de*  -f-  cd* 
consequently      -         ca*  +  ca*  =  ge*  -f  ce*  ; 
or,  by  doubling,  ab*  -j-  ab*  =  eg*  -j-  eg*. 


+  de*. 


-f  de* 


THEOREM  XU. 

The  difference  between  the  Semi-transverse  and  a  Line  drawn 
from  the  Focus  to  any  point  in  the  Curve,  is  equal  to  a 
Fourth  Proportional  to  the  Semi-transverse,  the  Distance 
from  the  Centre  to  the  Focus,  and  Distance  from  the 
Centre  to  the  Ordinate  belonging  to  that  Point  of  the 
Curve. 


T©fc.  I, 


62 


That 


48S 


eONIC  SECTIONS. 


That  is, 

AC  FE  =  CI,  or  FE  =  AI  ; 

and  /e  —  AC  =  ci,  of/e  =  m. 
Where  CA  i  cf  :  :  cd  :  ci  the  4th 
proportional  to  ca,  cf,  cd. 


For,  draw  ag  parallel  and  equal  to  ca  the  semi-conjugate  j 
and  join  cg  meeting  the  ordinate  de  in  h 
Then,  by  theor.  2  ca^  :  ag^  :  :  ca*  —  cd*  :  de*  : 
and,  by  sim.  tri.       oa*  :  ag*  ;  :  ca*  -—  cd*  :  ag*  —  dh*  ; 
consequently         de*  =  ag*  —  dh*  =  ca*  — dh*. 
Also  Fb  =  CF   cc  CD.  and  fd*  =  cf*  —  2gf  .  cd  -f-  cd*  ; 
but  by  right-angled  triangles,  fd*  -{-  de*  =  fe*  ; 
therefore  fe*  =  cf*  -f-  ca*  —  2cf  .  cd  -j-  cd*  —  dh*. 
But  by  theor.  4,         ca*  -f-  cf  =  ca*  ; 
and,  by  supposition,      2cf  .  cd  =  2ca  .  ex  ; 
theref.  fe*  =  ca*  —  2ca  .  ci  -f-  cd*  —  dh*  ; 
But  b}'  supposition,  ca*  :  cd*  :  :  cp^  or  ca*  — ^  ag*  :  ci*  ; 
and,  by  sim.  tri.        ca*  :  cd*  :  :  ca*  —  ag*    :  cd*  —  dh*  j 
therefore      -       -     ci*  =  cd*  —  dh*  ; 
consequently        -     fe*  =  ca*  —  2ca  .  ci  4-  ci*. 
And  the  root  or  side  of  this  square  is  fe  =  ca  —  ci  =  ai. 
In  the  same  manner  is  found^k:  =  ca  -|-  ci  =  bi.         q.  e.  d, 

,,  THEOREM  XIII. 

If  a  Line  be  drawn  from  either  Focus,  Perpendicular  to  a 
Tangent  to  any  I'oint  of  the  curve  ;  the  Distance  of  their 
Intersection  from  the  Centre  will  be  equal  to  the  Semi- 
transverse  Axis. 


That  is,  if  fp,  fp 
be  perpendicular  to 
the  tangent  Tpp, 
then  shall  cp  and 
cp  be  each  equal  to 


OP  THE  ELLIPSE.  483 

For,  through  the  point  of  contact  e  draw  pe,  and  /e  meet- 
ing:, FP  produced  in  g.  Then,  the  ^gkp  =  ^Ifep,  being  each 
equal  to  the  Z/k^»  and  the  angles  at  p  being  I'ight,  and  the  side 
PE  being  common,  the  two  triangles  gep,  fep  are  equal  in  all 
respects,  and  so  ge  =  fe,  and  gp  =  fp.  Therefore,  since 
Fp  =  ipG,  and  Fc  =  ^f/,  and  the  angle  at  f  common,  the  side 
cp  will  be  =  ^fo  or  ^ab,  that  is  cp  =  ca  or  cb.  And  in  the 
same  manner  cp  =  ca  or  cb.  q.  e.  d. 

Carol.   1.     A  circle  described  on  the  transverse  axis,  as  a 
diameter,  will  pass  through  the  points  p,  p  ;  because  all  the 
lines  cA,  cp,  cjo,  cb,  being  equal,  will  be  radii  of  the  circle. 
Corol    2.     cp  is  parallel  to/fi,  and  cp  parallel  to  fe. 

Corol.  3.  If  at  the  intersections  of  any  tangent,  with  the 
circumscribed  circle,  perpendicular  to  the  tangent  be  drawn, 
they  will  meet  the  transverse  axis  in  the  two  foci.  That  is, 
the  perpendiculars  pf  ,/j/'give  the  foci  f,/. 


THEOREM  XIV. 


The  equal  Ordinates,  or  the  Ordinates  at  equal  Distances 
from  the  Centre,  od  the  opposite  Sides  and  Ends  of  an 
Ellipse,  have  their  Extremities  connected  by  one  Right 
Line  passing  through  the  Centre,  and  that  Line  is  bisected 
by  the  Centre. 

That  is,  if  CD  =  cg,  or  the  ordinate  de  =  gh  ; 

then  shall  ce  =  ch,  and  ech  will  be  a  right  line  # 


For  when  cd  =  cg,  then  also  is  de  =  gh  by  th.  1. 
But  the  /_D  =  Zg,  bdng  both  right  angles  ; 
therefore  the  third  side  ce  =  ch,  and  the  ^dce  ==  ^gch, 
and  consequently  ech  is  a  right  line. 

Corol 


484 


CONIC  SECTIONS. 


CoroL  1.  And,  conversely,  if  ech  be  a  right  line  passing 
through  the  centre  ;  then  shall  it  be  bisected  by  the  centre, 
or  have  ce  =  ch  ;  also  de  will  be  =  gh,  and  cd  =  cg. 
.  Corol.  2.  Hence  also,  if  two  tang;ents  be  drawn  to  the  two 
ends,  E,  H  ©f  any  diameter  eh  ;  they  will  be  parallel  to  each 
other,  and  will  cut  the  axis  at  equal  angles,  and  at  equal  dis- 
tances from  the  centre  For,  the  two  cd,  ca  being  equal  to 
the  two  CG,  CB,  the  third  proportionals  ct,  cs  will  be  equal 
also  ;  then  the  two  sides  ce,  ct  being  equal  to  the  two  ch, 
■cs,  and  the  included  angle  ect  equal  to  the  included  angle  hcs,. 
all  the  other  corresponding  parts  are  equal  :  and  so  the 
^T=  ^s,  andiE  parallel  to  hs. 

Corol  3.  And  hence  the  four  tangents,  at  the  four  extrem- 
ities of  any  two  conjugate  diameters  form  a  parallelogram 
circumscribing  the  ellipse,  and  the  pairs  of  opposite  sides  are 
each  equal  to  the  corresponding  paralleL.conjugate  diameters. 
IPor,  if  the  diameter  eh  be  drawn  parallel  to  the  tangent  te 
or  HS,  it  will  be  the  conjugate  to  eh  by  the  definition  ;  and  the 
tangents  to  e,  h  will  be  parallel  to  each  other,  and  to  the  diam- 
eter EH  for  the  same  reason. 

THEOREM  XV. 

If  two  Ordinates  ed,  ed  be  drawn  from  the  Extremities  e,  e 
of  two  Conjugate  Diameters,  and  Tangents  be  drawn  to 
the  same  Extremities,  and  meeting  the  Axis  produced  in  t 
and  R  ; 

Then  shall  cd  be  a  mean  proportional  between  cd,  dR, 
and  cd  a  mean  proportional  between  c©,  dt. 


For,  by  theor.  7, 
and  by  the  same, 
theref.  by  equality, 
But  by  sim.  tri. 
theref.  by  equality, 
In  like  manner, 


CD 

CA 

:  c  A 

CT, 

cd 

CA 

:  CA 

CR 

CD 

cd 

:  CR 

CT, 

DT 

cd 

:  CT 

CR 

CD 

cd 

:  cd 

DT. 

cd 

►  CD 

:   CD 

dK, 

<l.  t. 


Corel 


or  THE  ELLIPSE. 


485 


ct>E  =  A  cde. 


dc 

dB, 

CA 

CT, 

AD 

AT, 

AD 

DT, 

DC 

DB. 

Carol.   1.     Hence  cd  :  cd  :  :  ck  :  ct. 
Carol.  2.     Hence  also  cd  :  cd  : ;  de  :  db. 
And  the  rectangle  cd  .  de  =  cd  .  de,  or  A 
Coral,  3.     Also  cd^  =  cd  .  dt, 
and  cd^  =  cd  .  dft. 
Or  cd  a  mean  proportional  between  cd,  dt  ; 
and  CD  a  mean  proportional  between  cd,  da. 

THEOREM  XVL 

The  same  Figure  being  constructed  as  in  the  last  Theorem 
each  Ordinate  will  divide  the  Axis,  and  the  Semi-axis  added 
to  the  external  Fart,  in  the  same  Ratio. 
[See  the  last  fig.] 
That  is,  DA  :  DT  :  :  DC  :  DB, 
and  dA  :  dR  :  : 
For,  by  theor.  7,  cd  :  ca  :  : 
and  by  div.  cd  :  ca  :  : 

and  by  comp.  cd  :  db  :  : 
or,  -  -  -  -  DA  :  DT  :  : 
In  like  manner,  dA  :  da  :  :  dc  :  do.  ^.  e.  d. 

Coral.   1.     Hence,  and  from  cor.  3  to  the  last,  it  is, 

c82    =  CD  .   dt  =  ad  .    DB  =  CA^— CD2, 

CD^  =  cd  .  dR  =  Ad  .  dfi  =  ca2  —cd^. 
Carol.  2.     Hence  also,  ca^  =r  cd^  -f  cd^, 
and  ca2  =  de^  -f  de^. 

Coral.  3.     Further,  because  ca^  :  ca^  : :  ad  .  db  or  cd^  ;  de^, 
therefore  ca    :  ca    :  :  cd  :  de. 
likewise    ca    :  ca    :  :  cd  :  de. 

THEOREM  XVn. 

If  from  any  Point  in  the  Curve  there  be  drawn  an  Ordinate, 
and  a  Perpendicular  to  the  Curve,  or  to  the  Tangent  at 
that  point :  Then,  the  ^ 

Dist.  on  the  Trans,  between  the  Centre  and  Ordinate,  cd  : 

Will  be  to  the  Dist.   pd  :  :  ^^ 

As  Sq  of  the  Trans.  Axis  :  ^ 

To  Sq.  of  the  Conjugate. 

That  is,  X 

CA^  :  ca'  :  :  DC  :  dp. 


486 


GONIC  SECTIONS. 


For,  by  theor.  2,  ca^  :  ca^  :  :  ad  .  db  :  de^  , 

But,  by  rt.  angled  As,  the  rect.  td  -  up  =  de^  ; 

and,  b}^  cor.  1,  theor.  16.  cd  .  dt  =  ad     db  ; 

therefore     -     -     ca^  :  ca^  :  :  td  .  dc  :  td  .  dp, 

or     -     -     -     -     ac2  :  ca2  : :  dc  :  dp.  q.  e.  d. 


THEOREM  XVIII. 


li  there  be  Two  Tangents  (Jrawn,  the  One  to  the  Extremity 
of  the  Transverse,  and  the  other  to  the  Extremity  of  any 
other  Diameter,  each  meeting  the  other's  diameter  pro- 
duced ;  the  two  Tangential  Triangles  so  formed  will  be 
equal. 


That  is, 
the  triangle  get  =  the 
triangle  can. 


For,  draw  the  ordinate  de.     Then 
By  sim.  triangles,  cd  :  ca  :  :  ce  :  cn  ; 
but,  by  theor.  7,    cd  :  ca  :  ;  ca  :  ct  ; 
theref.  by  equal,    ca  :  ct  :  :  ce  :  cn. 

The  two  triangles  cet,  can  have  then  the  angle  e  common, 
and  the  sides  about  that  angle  reciprocally  proportional  ;  those 
triangles  are  therefore  equal,  namely,  the  A  cet  =  A  can. 

Corol.  1.     From  each  of  the  equal  tri.  cet,  can, 
take  the  common  space  cape,  !; 

«     and  there  remains  the  external  A  pat  =:  A  pke. 

Corol.  2.     Also  from  the  equal  triangles  cet,  can, 
take  the  common  triangle  ced, 

and  there  remains  the  A  ted  =  trapez.  aned. 

THEOREM  XIX. 


The  same  being  supposed  as  in  the  last  Proposition  ;  then 
any  Lines  kq.  qg  drawn  parallel  to  the  two  Tangents,  shall 
also  cut  off  equal  Spaced.     That  is, 

Akqg 


OF  THE  ELLIPSE 

3sr 


4«7 


Akqg  =  trapez.  anhg, 
and  A  ity^  =trapez.  Atthg. 


For  draw  the  ordinate  de.     Then 
The  three  sim   triangles  can,  cde,  cgh, 
are  to  each  other  as  ca^,  cd^,  cg  ; 

th.  hy  div  the  trap,  aned  :  trap,  anhg  : :  ca^  —  cds  :  ca*  —  cg*. 
But,  by  theor.  1,        de^  :  gq^  ;  .  ca2-.cd2  cca^  — cg^, 

theref.  by  equ.  trap,  aned  :  trap,  anhg  :  :  de^  :  oq,^. 

but,  by  sim.  As,  tri.  ted  :  tri.     k€ig    :  :  de^  :  gq^  j 

theref.  by  equahty,  aned  :  ted    :  :  anhg         :  kqg. 

But,  by  cor.  2,  theor.  18,  the  trap,  aned  =  A  ted  ; 
and  therefore  the  trap  anhg  =  A  kqg. 
In  like  manner  the  trap.  Anhg  =  A  Ky^f.     q.  e.  d. 
Corol.  1.  The  three  spaces  anhg,  tehg,  kqg  are  all  equal. 
Corol.  2.    From  the  equals    anhg,  kqg, 
take  the  equals      AN/ig,  Kqg^ 
and  tliere  remains  ghna  =  gq^G» 

Corol.  3.    And  from  the  equals  ghna^  g^QQ, 
'  take  the  common  space  g^LHo, 
and  there  remains  the  A  L€tH  =  A  Lqh, 

Corol,  4.     Again  from  the  equals  kqg,  tehg, 
take  the  common  space  klhg, 
and  there  remains         telk  ==  A  lqh. 

Corol.  5.  Aiad  when, 
by  the  lines  kq,  gh, 
moving  with  a  parallel 
motion,  kq  comes  into 
the  position  ir,  where 
CR  is  the  conjugate  to 
CA  ;  then 

the  triangle      kqg  becomes  the  triangle  irc, 
and  the  space  anhg  becomes  the  triangle  anc  ; 
and  therefore  the  A  iRc  =  A  anc  =  A  tec. 

Corol.  6.    Also    when  the  lines   k^    and   nq,   by  moving 
with  a  parallel  motion,  come  into  the  position  ce,  Me, 

the 


488 


eONIC  SECTIONS; 


the  triangle     lqh  becomes  the  triangle  c«m, 

and  the  space  telk  becomes  the  triangle  tec  ; 

and  theref.  the  A  ccm  =  A  tec  =  a  anc  =  A  iRc. 


THEOREM  XX. 


Any  Diameter  bisects  all  its  Double  Ordinates,  or  the  Line« 
drawn  Parallel  to  the  Tangent  at  its  Vertex,  or  to  its  Con- 
jugate Diameter. 


That *s.  if  q,q  be  parallel  „ 
to  the  tangent  te,  or  to  cc,  -^ 
then  shall  lq  =  hq. 


For,  draw  qa,  qh  perpendicular  to  the  transverse. 
Then  by  cor.  3,  theor.  19,  the  A  lqh  =  A  J-qh ; 
but  these  triangles  are  also  equiangular  ; 
consequently  their  like  sides  are  equal,  or  lq  =■  t.q. 

Corol.  Any  diameter  divides  the  ellipse  into  two  equal 
parts. 

For,  the  ordinates  on  each  si4e  being  equal  to  each  other, 
and  equal  in  number  ;  all  the  ordinates,  or  the  area,  on  one 
side  of  the  diameter,  is  equal  to  all  the  ordiiiates,  or  the  area, 
on  the  other  side  of  it. 

THEOREM  XXI. 

As  the  Square  of  any  Diameter  : 
Is  to  the  Square  of  its  Conjugate  :  : 
So  is  the  Rectangle  of  any  two  Abscisses  : 
To  the  Square  of  their  Ordinate. 

That  is,  ce2  :  cc  :  :  el  .  lg  or  ce^  —  cl^  :  lq^  . 


For,  draw  the  tangent 
te,  and  produce  the  or- 
dinate QL  to  the  trans- 
verse at  K,  Also  draw 
^H,  CM  perpendicular 
to  the  transverse,  and 
meeting  eg  in  h  and  m. 

Then  similar  triangles 


being  as  the  squares  of  their  like  sides,  it  is, 


OF  THE  ELLIPSE. 


48d 


by  sim.  triangles,  A  get  :  A  clk  :  :  ce^  :  cl^  ; 

or,  by  divisioa,      A  c^t  '  trap,  telk  :  :  ce^  :  ce^  — cl^. 

Again,  by  sim.  tri.  A  ccm  ;  A  lqh  :  :  cc^  ;  lq,^. 
But,  by  cor.  5  theor.  19,  tbe  A  ccm  =  A  cet, 
and,  by  cor.  4  theor,  19,  the  A  lqh  =  trap,  telk  ; 
theref.  by  equality,  ce^  :  cc^  :  ;  ce^  — cl^  :  l^^^ 
or  .-        -         -  CE*  :  cc*  :  :  el  .  lg  :  l^^.     q.  e.  d. 

Corol.  1.  The  squares  of  the  ordinates  to  any  diameter, 
are  to  one  another  as  the  rectangles  of  their  respective 
abscisses,  or  as  the  difference  of  the  squares  of  the  semi- 
diameter  and  of  the  distance  between  the  ordinate  and  centre. 
For  they  are  all  in  the  same  ratio  of  ce^  to  ce^ . 

Corol.  2.  The  above  being  the  same  property  as  that  be- 
longing to  the  two  axes,  all  the  other,  properties  before  laid 
down,  for  the  axes,  may  be  understood  of  any  two  conjugate 
diameters  whatever,  using  only  the  oblique  ordinates  of  these 
diameters,  instead'of  the  perpendicular  ordinates  of  the  axes  ; 
namely,  all  the  properties  ii  theorems  6,  7,  8,  14,  15,  16, 
18  and  19. 

THEOREM  XXU. 


If  any  Two  Lines,  that  any  where  intersect  each  other,  meet 
the  Curve  each  in  Two  Points  ;  then 
The  Rectangle  of  the  Segments  of  the  one  : 
Is  to  the  Rectangle  of  the  Segments  of  the  other  :  : 
As  the  Square  of  the  Diam.  Parallel  to  the  former  : 
To  the  Square  of  the  Diam.  Parallel  to  the  latter. 


That  is,  if  or  and  cr,  be 
Parallel  to  any  two  Lines 
PHQ,  pnq  ;  then  shall 
Cft2  :  cr2  : :  PH  .  HQ  :  pH  .  ny. 


For,  draw  the  diameter  che,  and  the  tangent  te,  and  iti 
parallels  pk,  ri,  mh,  meeting  the  conjugate  of  the  diameter 
GR  in  the  points  t,  k,  i,  m.  Then,  because  similar  triangles 
are  as  the  squares  «f  their  like  sides,  it  is, 


VeL.  I. 


63 


bj 


490  CONIC  SECTIONS. 

by  sim.  triangles,     ch^  :  g?^  :  :  ^  cri  :  A  gfic, 

and       -       -       -       cr2  ;  gh^  :  :  /s,  chi  :  A  i-hm  ; 

theref  by  division,  cr^  :  gp^  —  gh^  ;  :  cri  :  kfhm. 

Again,  by  <iim.  tri.    ce^  :  ch^  :  :  A  cte  :  A  cmh  ; 

and  by  division,        ce^  :  ce^  ~—  ch^    :  :   A  cte  :  tehbi. 
But,  by  cor.  5  tbeor.  1 9,  the  A  cte  ^  A  cm, 
and  by  cor.'l  theor.  19,  tehc  =  kph».  or  tehm  =  kphm  ; 
theref.  by  equ,  ce^  :  ce^  — ch^  :  :  ch^  :  gp^  —  gh*  or  ph  .  hq,. 
In  like  manner  ce^  :  ce^  —  ch^  : :  cH  :  pu  .  Hg» 
Theref.  by  equ.  cr^  :  cr^  :  :  ph  .  uq.  :  pu  .  h^^  q.  e.  b. 

Corol.  1.  In  like  manner,  if  any  other  lines  p'b'^',  parallel 
to  cr  or  to  pq^  meet  phq  ;  ^Jince  the  rectangles  ph'q,  p'u^q' 
are  also  in  the  same  ratio  of  cr2  to  cr^  ;  therefore  rect. 
PHQ  :  puq  :  :  ph'q  :  p'u'^'. 

Also,  if  another  line  v'hq.'  be  drawn  parallel  to  pq  or  gr  : 
because  the  rectangles  p'/jq'  p'hq  are  still  in  the  same  ratio, 
therefore,  in  general,  the  rect.  phq,  :  puq  :  :  phq,'  :  phq'. 

That  is,  the  rectangles  of  the  parts  of  two  parallel  lines, 
are  to  one  another,  as  the  rectangles  of  the  parts  of  two  other 
parallel  lines,  any  where  intersecting  the  former. 

Corol.  2.  And  when  any  of  tbe  Jines  only  touch  the  curve, 
instead  of  cutting  it,  the  rectangles  of  such  become  squares, 
and  the  general  property  still  attends  them. 


That  is, 
cr2  :  cr^  :  :  te^  :  tc*  , 
or  CR  :  cr    :  :  te    :  ie. 
and  CR  :  cr  :  :  tE     :  te. 


Corol.  3.     And  hence  te  :  Te  :  :  tr,  :  te. 

OF 


[  491  ] 


OF  THE  HYPERBOLA. 

THEOREM  I. 

The. Squares  of  the  Ordinates  of  the  Axis  are  to  each  other 
as  the  Rectangles  of  their  Abscisses. 

Let  AVB  be  a  plape  passing 
through  the  vertex  and  axis  of 
the  opposite  cones  ;  a»jih  ano- 
ther section  of  them  perpendicu- 
lar to  the  plane  of  the  former  ; 
AB  the  axis  of  the  hyperbolic 
lections  ;  and  fg,  hi,  ordinales 
perpendicular  to  it.  Then  it  will 
ho  as  Fu^  :  hi^  : :  af,  fb  :  ah.hb. 

F'or,  through  the  oniinates  fg, 
HI,  draw  the  circular  sections 
K.  L,  Misr,  parallel  to  the  ba«<e  of 
.  the  cone,  having  kl,  m.v,  for  their 
diameters,  t;>  which  fg,  hi,  are  ordinates,  as  well  to  the  axis 
of  the  hyperbola. 

^Jow,    by   tile  similar  triangles  afl,  ahn,  and  bfk,  bhm, 
it  is  AF  :  AH  :  :  fl  :  hn, 
and  EB  :  UB  :  :  KF  :  MH ; 

hence,  takin»  the  rectangles   of  the  corresponding  terms, 

it  is,  the  rect.  af  .  fb  :  kh  .  hb  :  :  kf  .  fl  :  mh  ,  h.n. 
But,  by  the  cir:  l^,  kf  .  fl  =  fj^,  and  mh  .  hn  =  bi^  ; 
Therefore  the  feet,  af  .  fb  :  ah  .  hb  :  ;  fg^  :  hi^. 


THEOREM  H. 


As  the  Square  of  the  Transverse  Axis 
Is  to  the  Si|uare  of  the  Conjji;ate        : 
So  is  the  Rectangle  of  the  Abscisses 
To  the  Square  of  their  Ordinate. 


That  is,  ab«  :  ab^  or 

ag2  :  ac^  :  :  AD  .  db  :  de^. 


For, 


%9t 


CONIC  SECTIONS. 


For,  by  theor.  1 ,  ac  .  cb  :  ad  .  db  :  :  ca^  :  de^  ; 

But,  if  c  be  the  centre,  then  ac  .  cb  =  ac^,  and  ca  is  the 

semi-conj. 
Therefore         -         ac^  :  ad    .  db  :  :  ac^  :  de^  ; 
or,  by  permutation,       ac^  :  ac^  :  :  ad  .  db  :  de^  ; 
or,  by  doubling,  ab^  :  ab^  :  :  ad  .  db  :  de^  .     q.  e.  d. 

ab3 
CoroL     Or,  by  Aiv.  ab  :  —  ;  :  ad  .  db  or  cd^— ca^  :  de^, 

AB 

that  is,  AB  :  j9  :  :  AD  .  db  or  cd^  — ca^  :  de^  ; 
ab2 
where  p  is  the  parameter  —  by  the  definition  of  it^ 

AB 

That  is,  As  the  transverse, 
Is  to  its  parameter, 
So  ii?  the  rectangle  of  the  abscisses. 
To  the  square  of  their  ordinate. 

THEOREM  III. 


As  the  Square  of  the  Conjugate  Axis         : 

To  the  Square  of  the  Transverse  Axis   :  :. 

The  Sum  of  the  Squares  of  the  Semi-conjugate,  and 

"Distance  of  the  Centre  from  any  Ordinate  of  the  Axi^  \ 

The  Square  of  their  Ordinate. 


That  is, 
ca^  :  ca2  ;  :  ca2  +  cd2  dE^ . 


For,  draw  the  ordinate  ed  te  the  transverse  ab. 

Then,  by  theor.  1.  ca*  :  ca^  :  :  de^  :  ad  .  be  or  cd^  ^  ca^, 

or         -         -  ca2  :  ga^  :  :  cd^  :  dE^  —  ca^. 

But      r         -  ca2  :  ca3  :  :  ca2  :  ca2. 

theref.  by  compos,   ca^  :  ca^  :  :  ca«  +  cd^  :  ds^ . 

In  like  manner,        ca^  :  ca^  :  :  ca^  -f-  cd^  :  be^.       *.  e.  ». 

Carol.  By  the  last  theor.  ca^  :  ca2  :  :  cd3-— ca^  :  de*, 
and  by  this  theor.  ca^  :  ca^  :  :  cd2-|-ca  :  ^De^ 
therefore       -         de2  :  dc^  :  :  cd^— ca^  icD^-f-CA^. 
In  like  manner,  dea  :  dE*  : :  cd^— ca^  cd^  :  -fca*. 

theorem; 


OF  THE  HYPERBOLA. 
TIffiOREM  IV. 


495 


The  Square  of  the  Distance  of  the  Focus,  from  the  Centre,  ia 
equal  to  the  Sum  of  the  Squares  of  the  Semi  axes. 

Or,  the  Square  of  the  Distance  between  the  Foci,  is  equal  to 
the  Slim  of  the  Squares  of  the  two  Akcs. 


That  is, 

cp2  =  cas  -|-  ca3,  or 
Ffa    =  ab3  -f.  ab3 


For,  to  the  focus  f  draw  the  ordinate  fe  ;  which,  by  the 
definition,  will  be  the  semi-parameter.  Then,  by  the  nature 
of  the  curve  -  ca^  :  ca^  :  :  cf^  —  ca^  :  fe^, 

and  by  the  def.  of  the  para,  ca^  :  ca^  :  :         ca^        :  fe^  ; 
therefore  -         -         ca^  =  cf^  —  ca^  j 

and  by  addition,  -         cf^  =  ca^  -}-  ca^  ; 

or,  by  doubling,  -         rf^   =AB2  4-ab2.  q,  e.  Do 

Corol.  1.  The  two  semi-axes,  and  the  focal  distance  from 
the  centre,  are  the  sides  of  a  right-angled  triangle  CAa  ;  and 
the  distance  Aa  is  =  cf  the  focal  distance. 

Corol.  2.  The  conjugate  semi-axes  ca  is  a  mean  propor- 
tional between  af,  fb,  or  between  Af,  fs,  the  distances  of  either 
focus  from  the  two  vertices. 

For  ca2  =  cf^  —  ca^  =  (cf  -f-  ca)  .  (cp  —  ca)  =  af  .  fb- 

THEOREM  V. 

The  Difference  of  two  Lines  drawn  from  the  two  Foci,  to 
meet  at  any  Point  in  the  Curve,  is  equal  to  the  Transverse 
Axis. 


That  H, 

fe FE  =  AB. 


For,  draw  ag  parallel  and  equal  to  ca  the  semi-conjugate  ; 
and  join  cg,  meeting  the  ordinate  de  produced  in  h  :  also  take 
ci  a  4th  proportional  to  ca,  cf,  cb. 

Then, 


494,  CONIC  SECTIONS. 

Then  by  th.  2,  ca^  :  ag^  :  :  cd2  —  ca^  :  de2  ; 
and,  by  sim.  As»  ca^  :  ag^  :  :  cd^  —  ca*  :  dh»  —  ag  j 
consequently         de^  =  dh2  — ag^  =  dh^  — ca*. 
Also,  FD  =  CF  i/:  CD,  and  fd^  =  cf^— 2cf  .  cd  -H  cd^  j 
and,  by  right-angled  triangles,  fe^  =  fd^  -f-  de^ 
therefore  fe^  =  cf2  —  ca^  —  2cf  .  cd  -f*  cd^  4"  dH^v 
But  by  theor.  4,         cf^  —cas  =  ca^^ 
and,  by  supposition,  2cf  .  c»  =  2ca     ci  ; 
theref.  fe^  =  ca2— .  2ca  .  ci -f  cd^  +  dh^  ; 
Again,  by  suppos.       ca^  :  cd^  :  ;  cf^  or  ca^  -{-  ag^  :  ci^  ; 
and,  by  sim.  tri.  ca^  :  cd^  :  :  ca^  -f- ag^  :  cd^  -f*  dh*  ; 

therefore  -         ci^  =  cd^  -f-  dh^  =  ch^  ; 

consequently  fe^  =  ca^    -  2ca  .  ci,+  ci*. 

And  the  root  or  side  of  tl  is  square  is  ff  =  ci  —  ca  =  ai. 
In  the  same  manner ,  it  it;  found  that  fE  =  ci  +  ca  =  bi. 
Conseq.  by  subtract,  fs  —  fe  =  bi  —  ai  =  ab.  q,.  e.  d. 

Corol.  1.     Hence  ch  =  ci  is  a  4th  proportional  to  ca,  of, 

CD. 

Corol.  2.  And  fs  -f-  fe  =  2ch  or  2ci ;  orFE,  ch,  fE,  are  in 
continued  arithmetrcal  progression,  the  common  diflference 
being  ca  the  semi-transverse. 

Corol.  3.  Hence  is  derived  the  common  method  of  describ- 
ing this  curve  mechanically  by  points,  thus  ; 

In  the  transverse  ab,  produced,  take  the  foci  f,  f,  and  any 
point  I.  Then  with  the  radii  ai,  bi,  and  centres  f,  f,  describe 
arcs  intersecting  in  e,  which  will  be  a  point  in  the  curve.  In 
like  manner,  assuming  other  points  i,  as  many  other  points  will 
be  found  in  the  curve. 

Then,  with  a  steady  hand  the  curve  line  may  be  drawn 
through  all  the  points  of  intersection  e. 

In  the  same  maner  are  constructed  the  other  two  or  conju- 
gate hyperbolas,  using  the  axis  ab  instead  of  ab. 

THEOREM  VI. 

If  from  any  Point  i  in  the  Axis,  a  line  il  be  drawn  touching 
the  Curve  in  one  point  l  ;  and  the  Ordinate  lu  he  drawn  : 
and  if  c  be  the  Centre  or  the  Middle  of  ab  :  Then  shall  cm 
be  to  ci  as  the  Square  of  am  to  the  Square  of  ai. 

H 

That  is,  y:^^\         \ 

CM  :  ci  :  :  am^  :  ai^  .  J^/Y"  ' 

"~ll        C    lA^DMKG 

For, 


OF  THE  HYPERBOLA.  496 

for,  from  the  point  i  draw  any  line  ieh  to  cut  the  curve  in 
two  points  E  and  h  ;  from  which  let  fall  the  perps.  ed,  hg  ;  and 
bisect  ou  in  K. 

Then  by  theor.  1,         ad  .  db  :  ag  .  gb  :  :  Dt^  :  gh^, 
and  by  sim.  triangles,  id^       :     ig^       :  :  de^  :  gh^  ; 

therof.  by  equality;  ad  .  db  :  ag     gb  :  :  id^   :  ig^  ; 

But  db  =  CB  H-  CD  =  CB  4-  CD  =  CG  -j-  CD  —  AG  =  2cK  —  AG, 

and  GB  =  4-  CG  =  cA  -f-  cg  =  cg  +  cd  —  ad  =  2ck  —  ad  ; 
theref  ad  .  2ck  —  ad  .  ag  :  ag  .  2ck— ad  .  ag  :  :  id^  :  ig^, 
and,  by  div»  dg  .  2ck  :  ig^  —  id^  or  dg  .  2ik  :  :  ad  .  2gk 

AD  .  AG  :  ID^ . 

or      -      2cK  :  2ik  :  :  ad  .  2ck  —  ad  .  ag  :  id^  ; 
or     AD  .  2cK  :  ad     2ik  :  :  ad  .  2ck--ad  .  ag  :  id^  ; 
theref.  by  div.  ck  :  ik  :  :  ad  ,  ag  :  ad  .  2ik  —  io^, 

and,  by  div.     ck  :  ci  :  :  ad  .  ag  :  id^  —  ad  .  id  -|-  ia, 
or         -  CK  :  CI  :  :  ad  .  ag  :  ai^. 

But,  when  the  line  m,  by  revolving  about  the  point, i, 
comes  into  the  position  of  the  tangent  il,  then  the  points  e 
and  H  meet  in  the  point  l,  and  the  points  d,  k,  g,  coincide 
with  the  point  m  ;  and  then  the  last  proportion  becomes  gm  : 
ci  :  :  am2  :  ai^.  q.  e.  d. 

THEOREM  VII.      " 

If  a  Tangent  and  Ordinate  be  drawn  from  any  Point  in  the 
Curve,  meeting  the  Transverse  Axis  ;  the  Semi-transverse 
V   will  he  a  Mean  Proportional  between  the  Distances  of  the 
jaid  Two  Intersections  from  the  centre. 

^B 
That  is,  \\  ^E 

CA  is  a  mean  proportional  between 
CD  and  CT  ;  or  cd,  ca,  ct,  are  con-         ■ — jg — ^ 
tinued  proportionals. 


\° 


^  For,  by  th.  6,  cd  :  ct  :  :  ad^  :  at^  , 
that  is,    -    CD  :  CT  :  :  (cd  — ca)^  :  (ca  — ct)^, 
or    -   -    CD  :  CT  :  :  cd^  -f*  ca^  :  ca^  -f  ct^, 
and   -   -    CD  :  DT  :  :  cd^  -f-  ca^  :  cd^  —  ct^, 
or    -   -    CD  :  DT  :  :  CD^  -f"  ca^  :  (cd  +  ct)  dt, 
or  '    CD^  :  CD  .  DT  :  :  cd^  -|-  ca^  :  cd  .  dt  -f-  ct  .  td  ; 
hence  cd^  :  ca^  :  :  cd  .  dt  :  ct  .  td, 
and    cd3  :  ca  :  :  cd  :  ct, 

theref.  (th.  78,  Geom.)  cd  :  ca  :  :  ca  :  cd.        *ft.  e.  d.~ 

Carol 


496  CONIC  SECTIONS. 

CoroL  Since  ct  is  always  a  third  proportional  to  cb,  ca  ; 
if  the  points  d,  a,  remain  constant,  then  will  the  point  t  be 
constant  also  ;  and  therefore  all  the  tangents  will  meet  in  thii 
point  T,  which  are  drawn  from  the  point  e,  of  every  hyper-  , 
bola  described  on  the  same  axis  ab,  where  they  are  cut  by  the 
common  ordinate  dfe  drawn  from  the  point  d. 

tueJorem  vm. 

If  there  be  any  Tangent  meeting  Four  Perpendiculars  to  the 
Axis  drawn  from  these  four  Points,  namely,  the  Centre,  the 
two  Extremities  of  the  Axis,  and  the  Point  of  Contact  ; 
those  Four  Perpendiculars  Will  be  Proportionals. 


That  is, 
AC  :  DE  :  :  CH  :  Bi. 

^  Tl 

For,  by  theor,  7,  to  :  ac  :  :  ac  :  do, 

theref.  by  div.  ta  :  ad  :  :  tc  ;  ac  or  cb, 

and  by  comp.  ta  :  td  :  :  tc  :  tb, 

and  by  sim.  tri.  ag  :  de  :  :  ch  :  bi.  q. 

Carol.  Hence  ta,  td,  tc,  tb,  ?  i  -a-      i 

^„  1  „  '  }  are  also  proportionals, 

and  tg,  te,  th,  ti,  ^  f    r 

For  these  are  as  ag,  de,  ch,  bi,  by  similar  triangles. 


THEOREM  IX. 

If  there  be  any  Tangent,  and  two  Lines  drawn  from  the  Foci 
to  the  Point  of  Contact ;  these  two  Lines  will  make  equal 
Angles  with  the  Tangent. 


That  is, 
the  Z,  fet  =  j^fEe, 


For,  draw  the  ordinate  de,  and  fe  parallel  to  ee. 
By  cor.  1 ,  theor.  6,  ca  :  cd  :  ;  cf  :  ca  -f-  fe, 
and  by  th.  7.  <;a  :  €»  :  ct  :  ca  ; 

therefore 


OF  THE  HYPERBOLA. 


497 


therefore         -  ct  :  cp  :  :  ca  :  ca  -|-  fe  ; 

and  by  add.  and  sub.  tp  :  xf  :  :  fe  :  2ca4-  fe  orfeby  th.  6. 
But  by  sim.  tri.  tf  :  xf  :  :  fe  :  tE  ; 

therefore  -         fe  =  fe,  and  conseq      ^e  =  ^fee. 

But,  because  fe  is  parallel  to  fe,  the  Ze  =  Z*»2T  ; 

tiierefore  the  ^fet  =  i^fEe.  q,.  e.  »• 

Corol.  As  opticians  find  that  the  angle  of  incidence  is  equal 
to  the  angle  of  reflexion,  it  appears,  from  this  proposition, 
that  ray«  of  light  issuing  from  the  one  focus,  and  meeting  the 
curve  in  every  point,  will  be  reflected  into  lines  drawn  from 
the  other  focus.  So  the  ray  fE  is  reflected  into  fe.  And  tliis 
is  the  reason  why  the  points  f,  f,  are  called  /oa,  or  burning 
points. 

THEOREM  X. 

All  the  Parallelograms  inscribed  between  the  four  Conjugate 
Hyperbolas  are  equal  to  one  another,  and  each  equal  to  the 
Rectangle  of  the  two  Axes. 


That  is, 
the  parallelogram  p^rs  = 
the  rectangle  ab  .  ab. 


Let  EG,  eg  be  two  cogjugate  diameters  parallel  to  the  sides 
of  the  parallelogram,  and  dividing  it  into  four  less  and  equal 
parallelograms.  Also,  draw  the  ordi«ites  de,  de,  and  ck 
perpendicular  to  pq  ;  and  let  the  axis  produced  meet  the  sides 
of  the  parallelograms,  produced,  if  necessary,  in  t  and  t. 
Then,  by  theor.  7,  ct  :  ca 


CA     :     CD, 

;  CA  :  cd  ; 

cd  :   CD  ; 

TD  :  cd, 

cd  :  CD, 


and  -         -         ct 

theref  by  equality      ex  :  ct 

but,  by  sim,  triangles,  ex  :  ct 

theref  by  equality,     xd  :  cd 

and  the  rectangle        xd  :  dc  is  =  the  square  cd^. 

Again,  by  theor.  7,     cd  ;  ca  :  :  ca  :  ex, 

or,  by  division,  cd  :  ca  :  :  da  :  ax, 

and,  by  composition,    cd  :  db   :  :  da  :  dx  ; 

eonseq  the  rectangle  cd  :  dx  =  cd^  =  ad  .  db*. 


*  Carol.  Because  cJ>  =  ad  .  db  =3  cd«  «^  CA>. 

therefore  ca*  =  cd^  —  cd^. 
In  like  manner  ca*  =s  de*  —  dez. 
Vol.  K  64 


But, 


4^^ 


CONIC  SECTIONS. 


But,  by  theor.  1, 

eA3 

:  ca3  :  :  (ad  .  db  or)  cd*  :  be», 

therefore 

CA 

:  ca    :  :  cd  :  DE  ; 

In  like  manner, 

CA 

:  ca    :  :  CD  :  de ; 

or            -          - 

ca 

:  de    :  :  ca  :  cd. 

But,  by  theor.  7, 

CT 

:  ca    t:  CA  :  cd  ; 

theref.  by  equality, 

CT 

:  CA    :  :  ca  :  de. 

But,  by  sim.  tri. 

CT 

:  CK    :  ;  ce  :  de  ; 

theref  by  equality, 

CK 

:  CA    :  :  ca  ;  ce. 

and  the  rectangle 

CK 

.  ce   r=  CA  .  ca. 

Byt  the  rect. 

CK 

.  ce  —  the  parallelogram  cepc, 

theref.  the  rect. 

CA 

.  ce  =  the  parallelogram  cepc, 

conseq.  the  rect. 

AB 

.  ab  =  the  paral.  p^rs.      q.  e. 

THEOREM  XL 

The  Diflference  of  the  Squares  of  every  Pair  of  Conjugate 
Diameters,  is  equal  to  the  same  constant  Quantity,  namely, 
the  Difference  of  the  Squares  of  the  two  Axes. 

That  is, 
ab2  —  ab2  =  eg2  —  eg3  ;  \   d 

where  eg,  eg  are  any  conjugate 
diameters. 

For,  draw  the  ordinates  ed,  ed. 
Then,  by  cor.  to  theor.  10,  ca^  =  cd^  —  cd^, 
and  -         -         -         -     ca2  =  de^  —  de^  ; 

theref.  the  difference  ca«  —  ca^  =  cd^  -f-  de*  — cd^  —  de*. 
But,  by  right-angled  As,  ce*  =  cd*  -{-  de*  ; 

and  -         -         -         -    ce*  =  cd*  -f  de*  ; 

theref.  the  difference  ce^  —  ce*  =  cd*  -{-  de*  —  cd*  —  de 
consequently        -         ca*  —  ca*  =  ce*  —  ce*  ; 
or,  by  doubling,     .        ab*  — ab^  =  eg*— eg*.         q.  e.  ». 

THEOREM  XU. 

All  the  Parallelograms  are  equal  which  are  formed  between 
the  Asymptotes  and  Curve,  by  Lines  drawn  Parallel  to  the 
Asymptotes. 

That  is,  the  lines  ge,  ek,  ap,  aq, 
being  parallel  to  the  asymptotes  ch,  cl  ; 
then  the  paral.  cgek  =  paral.  cpaq. 


CXQ 


OP  THE  HYPERBOLA.  499 

For,  let  A  be  the  vertex  of  the  curve,  or  extremity  of  the 
semi-transverse  axis  ac,  perp.  to  which  draw  al  or  a1,  which 
will  be  equal  to  the  serai-conjugate,  by  definition  19.  Also, 
draw  HEoeh  parallel  to  l1, 

Then,  by  th'eor.  2,  ca^    :  al*   ::  cd^-caS  :  m^, 
and,  by  parallels,         ca*   :  al*   ;  :  cd^  :  dh^  ; 
theref.  by  subtract,      cas   :  al«  z  :  ga^  :  dh?  —  de    or 

rect.  HG  .  Eh  ; 
coQseq.  the'square  ai:«  =  the  rect.  he  .  Eh. 

But,  by  sim.  tri.  pa  :  al  :  :  ge  :  eh, 
and,  by  the  same,  qa  :  a\  :  :  ek  :  Eh  ; 
theref.  by  comp.    pa   :  a^  :  al^    :  :  ge  -  ek  :  he  .  Eh  ; 
and,  because  al^  =  he   .  Eh,  theref.  pa      A(i=GE.EK. 

But  the  parallelograms  cgek,  cpaq,  being  equiangular,  are 
as  the  rectangles  ge  .  ek  and  pa  .  aq. 

Therefore  the  parallelogram  gk  =  the  paral.  pq. 

That  is,  all  the  inscribed  parallelograms  are  equal  to  one 
another.  Q-  e.  d. 

Corol.  1.  Because  the  rectangle  gek  or  cge  is  constant, 
therefore  ge  is  reciprocally  as  cg,  or  cg  :  cp  :  :  pa  :  ge. 
And  hence  the  asymptote  continually  approaches  towards  the 
curve,  but  never  meets  it :  for  ge  decreases  continually  as 
CG  increases  ;  and  it  is  always  of  some  magnitude,  except 
when  cG  is  supposed  to  be  infinitely  great,  for  then  ge  is 
infinitely  small,  or  nothing.  So  that  the  asymptote  cg  may  be 
considered  as  a  tangent  to  the  curve  at  a  point  infinitely  distant 
fromc. 

Corol.  2.  If  the  abscisses  cd,  ce, 
CG,  &c.  taken  on  the  one  asymp- 
tote, be  in  geometrical  progression 
increalsing  ;  then  shall  the  ordi- 
nates  dh,  di,  gk,  &c    parallel  to        /  ^'^'^^^L.^I 

the  other  asymptote,  be  a  decreas- 
ing geometrical  progression,  hav-  ^ 
ing  the  same  ratio.  For,  all  the 
rectangles,  cdh,  cei,  cgk,  kc.  being  equal,  the  ordinates  dh, 
EI,  gk,  &c.  are  reciprocally  as  the  abscisses,  cd,  ce,  cg,  kc» 
which  are  geometricals.  And  the  reciprocals  of  geornetricals 
are  also  geometricals,  and  in  the  same  ratio,  but  decreasing,  or 
in  converse  order. 


THEOREM 


50t  CONIC  SECTIONS. 


THEQREM  XIII. 

The  three  following  Spaces,  between  the  Asymptotes  and 
the  Curve,  are  equal  ;  namely,  the  Sector  or  'J  rihnear 
Space  contained  by  an  Arc  of  the  Curve  and  two  Radii, 
or  Lines  drawn  from  its  Extremities  to  the  Centre  ;  and 
each  of  the  two  (Quadrilaterals,  contained  by  the  said  Arc, 
and  two  Lines  drawn  from  its  Extremities  parallel  to  one 
Asymptote,  and  the  intercepted  Part  of  the  other  Asymp- 
tote. 


That  is. 
The  sector  cae  =  paeg  ==  qaek, 
all  standing  on  the  same  arc  ae. 


For,  by  theor.  12,  cpaq  =  cgek  ; 
subtract  the  conimon  space  CGiq, 
there  remains  the  paral.  pi  =  the  par.  ik  ; 
to  each  add  the  trilineal  iae,  then 
the  sum  is  the  quadr.  paeg  =  qaek. 

Again,  from  the  quadrilateral  caek 
take  the  equal  triangles  caq,  cek, 
and  there  remains  the  sector  cae  =  qaek. 
Therefore  cae  =  qaek  =  paeg.  ^.  e.  ©,. 

THEOREM  XIV. 

The  Sum  or  Difference  of  the  Semi-transverse  and  a  Line 
drawn  from  the  Focus  to  any  1  oint  in  the  Curve,  is  equal 
to  a  Fourth  Proportional  to  the  Semi-transverse,  the  Dis- 
tance from  the  Centre  to  the  Focus,  and  the  distance  from 
the  Centre  to  the  Ordinate  belonging  to  that  Point  of  the 
Curve. 


That  is, 
FE  -}-  AC  =  CI,  or  fe  ==  ai; 
and  /e  —  AC  =  ci,  or  /k  =  bi. 
Where  ca  :  cf  :  :  cd  :  ci  the 
4th  propor.  to  ca,  cf,  cd. 


OF  THE  HYPERBOLA.  601 

Pop,  draw  ag  parallel  and  equal  to  ca  the  semi-conjugate  ; 
and  join  cc  meeting  the  ordinate  de  produced  in  h. 
Then,  by  theor.  2,  ca^  :  ag2  :  :  cd^  — ca^  :  de2  ; 
and,   by  sim.   As,  ca^  :  ag^  :  :  cd^  — ca^  :  dh^  —  ag^  ; 
consequently  de^  =  dh-  —  ag^  =  dh^  —  ca^. 

Also   FD  =  CF  GO    CD,  and  fd^  =  cf^  —  2cf  .  cd  -f  cd^  ; 
but,  by  right  angled  triangles,  fd^  -\-  de^  ==  fe^  ; 
therefore  fe^  =cf2  —  ca^  ,—  2cf  .  cd  -|-  cd^  -j-  dh^. 

But  by  theor.  4,         cf^  —  ca^  =  ca^, 
and,  by  supposition,       2cf  .  cd  =  2ca  .  ci ; 
theref.  fe^  =ca2  —  2ca  .  ci  -{-  cd^  -|-  dh^. 

But,  by  supposition,  ca^  :  cd^  :  :  cf^  or  ca^  -\-  ag^  :  ci^  ; 
and,  by  sim.  As,  ca^  :  cd^  :  :  ca^  -f-  ag^  :  cd^  -|-  dh^  ; 
therefore  -  ci^  =cd2  -|-  dh^  =ch2  ; 

consequently      -      fe2=ca2  —  2ca  .  ci  +  ci^. 
And  the  root  or  side  of  this  square  is  fe  =ci  —  ca  =  ai. 
In  the  same  manner  is  found /e  =  ci  -|-  ca  =  bi.  q.  e.  d. 

Corol.  From  the  demonstration  it  appears,  that  de^  = 
dh^  —  Ao2  =  dh^  —  ca.  Consequently  dh  is  every  where 
greater  than  de  ;  and  so  the  asymptote  cgh  never  meets 
the  curve,  though  they  be  ever  so  far  produced  :  but  dh  and 
de  approach  nearer  and  nearer  to  a  ratio  of  equality  as  they 
recede  farther  from  the  vertex,  till  at  an  infinite  distance  they 
become  equal,  and  the  asymptote  is  a  tangent  to  the  curve  at 
an  infinite  distance  from  the  vertex. 


THEOREM  XV. 

If  a  Line  be  drawn  from  either  Focus  Perpendicular  to  a 
Tangent  to  any  Point  of  the  curve  ;  the  Distance  of  their 
Intersection  from  the  Centre  will  be  equal  to  the, Semi- 
transverse  Axis. 


That  is,  if  ep,  fp  be  perpen- 
flicular  to  the  tangent  tp/),  then 
shall  cp  and  cp  be  each  equal 

to  CA  or  CB. 


602  CONie  SECTIONS. 

For,  through  the  point  of  contact  e  draw  fe  and  /e,  meet- 
ing FP  produced  in  g.  Then,  the  ^gep  =  ^kep,  being  each 
equal  to  the  Z/Ep,  and  the  angles  at  p  being  right,  and  the 
side  PE  being  common,  the  two  triangles  gep,  fep  are  equal 
in  all  respects,  and  so  ge  =  fe,  and  gp  =  fp.  Therefore, 
since  fp  =iFG,  and  fc  =  1f/,  and  the  angle  at  f  common, 
the  side  cp  will  be  =  ^/g  or  \ab.  that  is  cp  =  ca  or  cb.  ,, 

And  in  the  same  manner  cp  =  ca  of.cb.  q.  e.  d.  * 

Corol.  1.  A  circle  described  on  the  transverse  axis  as  a 
diameter,  will  pass  through  the  points  p,  p  ;  because  all  the 
lines,  CA,  cp,  cp,  cb,  being  equal,  will  be  radii  of  the  circle. 

Corol.  2.  cp  is  parallel  to/E,  and  cp  parallel  to  fe. 

Corol.  3.  If  at  the  intersections  of  any  tangent,  with  the 
circumscribed  citcle,  perpendiculars  to  the  tangent  be  drawn, 
they  will  meet  the  transverse  axis  in  the  two  foci.  That  is, 
the  perpendiculars  pf,  pfgive  the  foci  f,/. 


THEOREM  XVI. 

The  equal  Ordinates,  or  the  Ordinates  at  equal  Distances 
fr.  m  the  Centre,  on  the  opposite  Sides  and  Ends  of  an 
Hyperbola,  have  their  Extremities  connected  by  one  Right 
Line  passing  through  the  Centre,  and  that  Line  is  bisected 
by  the  Centre. 


That  is,  if  cd  =  cg,  or  the 
ordinate  dk  =  gh  ;  then  shall 
CE  =  cH,  and  ech  will  be  a 
right  line. 


For,  when  cd  =  cg,  then  also  is  de  ==  gh  by  cor.  2  theor.  1, 
But  the  Zd  =  Zg,  being  both  right  angles  ; 
therefore  the  third  side  ce,=  ch,  and  the  Zdce  =  Z^CH, 
and  consequently  ech  is  a  right  line. 

Corol.  1.  And,  conversely,  if  ech  be  a  right  line  passing 
through  the  centre  ;  then  shall  it  be  bisected  by  the  centre, 
or  have  ce  =  ch  ;  also  de  will  be  =  gh,  and  cd  =  cg. 

Corol.  2.  Hence  also,  if  two  tangents  be  drawn  to  the  two 
ends  e,  h  of  any  diameter  eh  ;  they  will  be  parallel  to  each 

other, 


OF  THE  HYPERBOLA. 


503 


other,  and  will  cut  the  axis  at  equal  angleS)  and  at  equal  dis- 
tances from  the  centre.  For,  the  two  cd,  ca  being  equal  to 
the  two  CG,  CB,  the  third  proportionals  ct,'  cs  will  be  equal 
also  ;  then  the  two  sides  ce,  ct  being  equal  to  the  two  ch,  cs, 
and  the  included  angle  ect  equal  to  the  included  angle  hcs, 
all  the  other  corresponding  parts  are  equal:  and  so  the  ^t 
.  =  ^s,  and  TE  parallel  to  hs. 

Corol.  3.  And  hence  the  four  tangents,  at  the  four  ex- 
tremities of  any  twe  conjugate  diameters,  form  a  parallelogram 
inscribed  between  the  hyperbolas,  and  the  pairs  of  opposite 
sides  are  each  equal  to  the  corresponding  parallel  conjugate 
diameters. — For,  if  the  diameter  eh  be  drawn  parallel  to  the 
tangent  te  or  hs,  it  will  be  the  conjugate  to  eh  by  the  defi- 
nition ;  and  the  tangents  to  eh  will  be  parallel  to  each  other, 
and  to  the  diameter  eh  for  the  same  reason. 

THEOREM  XVn. 

If  two  Ordinates  ED,  ed  be  drawn  from  the  Extremities  e,  c, 
of  two  Conjugate  Diameters,  and  Tangents  be  drawn  to  the 
same  Extremities,  and  meeting  the  Axis  produced  in  t 
and  R  ; 

Then  shall  cd  be  a  mean  Proportional    between  cd^  Jr, 
and  cd  a  mean  Proportional  between  cd,  dt. 


For,  by  theer.  7,       cd  :  ca  : :  ca  :  ct, 
■    and  by  the  same,       cd!  :  ca  :  :  ca  :  cr  j 

theref.  by  equahty,    cd  :  cd  :  :  cK  :  ct. 

But  by  sim.  tri.  dt  :  cd  :  :  ct  :  cr  ; 

the  ef.  by  equality,  cd  :  cd  : :  cd  :  dt. 

In  like  manner,  cd  :  cb  :  :  cd  :  dK. 

Corol.   1,     Hence  cd  :  cd  :  :  CK  :  ct,. 
Corol.  2.     Hence  also  cd  :  cd  : :  de  :  de. 
and  the  XQct.  cd  .  de  =  cd  .  de,  or  A  cde  =  A  cde. 
Corol,  3.     Also  cd^  =  cd  .  dt,  and  cd^  =  cd  .  dK. 

Or  cd  a  mean  proportional  between  cd,  dt  ; 

and  CD  a  mean  proportional  between  cdy  dn. 


Q.  £.  d. 


THEOREM 


504  CONIC  SECTION^^. 


THEOREM  XVIIL 

The  same  Figure  being  constructed  as  in  the  last  Proposition, 
each  Ordinate  will  divide  the  Axis,  and  the  Semi-axis  add- 
ed to  the  external  Part  j^n  the  same  Ratio. 


[See  the  last  fig.] 


That  is,  DA 

:  DT  :  :  dc 

:  DB, 

and  ^A 

:  dR  :  :  dc 

:  dB. 

For,  by  theor.  7, 

CD  :  CA   :  : 

CA  : 

CT, 

and  by  div. 

CD  :  CA   :  : 

AD  : 

AT, 

and  by  comp. 

CD  :  DB    :  : 

AD 

DT, 

or 

DA  :'  DT  :  : 

DC 

DB. 

In  like  manner, 

(Ia  :  dR  :  : 

dc 

dB. 

Corol.   1.     Hence,  and  from  cor.  3  to  the  last  prop,  it  is, 

cd^    =  CD  .  DT  =  AD  .  DB  =  CD^    CA^ , 

and  cd  .  dR  =  Ad  .  ds  =  ca^  — cd^ . 
Corol.  2.  Hence  also  ca^  =  cd^  —  cri^^andca^  =c?e2— de^. 
Corol.  3.     Farther,  because  ca^  :  ca^  :  :  ad  .  db  or  cd^  :  be^. 
therefore  ca  :  ca  :  :  cd  :  de. 
likewise    ca  :  ca  :  :  cd  :  de. 


THEOREM  XIX. 

If  from  any  Point  in  the  Curve  there  be  drawn  an  Ordinate, 
and  a  Perpendicular  to  the  Curve,  or  to  the  Tangent  at  that 
Point:  Then  the 

Dist.  on  the  Trans,  between  the  Centre  and  Ordinate,  cd  : 

Will  be  to  the  Dist.  pa  :  : 

As  Square  of  Trans.  Axis  : 

To  Square  of  the  Conjugate. 

That  is, 
CA^  :  ca^  :  :  DC  :  dp. 


For,  by  theor.  2,  ca^  :  ca^  :  :  ad  .  db  :  de^  , 

But,  by  rt.  angled  As?  the  rect.  t»  .  bp  =  de^  ; 

and,  by  cor.  1  theor.  16,  cd  .  dt  =  ad  .  db  ; 

therefore     -     -     ca^  :  ca^  :  :  td  .  dc  :  td  .  dp, 

or      -     -     .     _     cA^  :  ca^  :  :  do  :  dp  q.  e.  d. 

THEOREM 


OF  THE  HYPERBOLA. 


605 


TUCO^EM  XX. 


ff  there  be  Two  Tangents  drawn,  the  One  to  the  Extremity 
of  the  Transverse,  and  the  other  to  the  Extremity  of  any 
other  Diameter,  each  meeting  the  other's  Diameter  pro- 
duced ;  the  two  Tangential  Tri?ingles  so  formed,  will  he 
equal. 


That  is, 
the  triangle  get  = 
the  triangle  can 


For,  draw  the  ordinate  de.     Then 
By  sim.  triangles,  cd  :  ca  : :  ce  :  cn  ; 
bat,  by  theor.  7,  cd  :  ca  :  :  ca  :  ct  ; 
theref.  by  equal,  ca  :  ct  :  :  ce  :  cn. 
The  two  triansjles  cet,  cav  have  then  the  angle  c  common, 
and  the  fiides  about  that  angle  reciprocally  proportional :  those 
triangles  are  therefore  equal,  viz.  the  A  cet  =  can.  q.  k.  b. 
Corol.   1.     Take  each  of  the  equal  tri.  qkt,  can,    . 
from  ttie  common  space  cape, 
and -there  remains  the  external  A  pat  =  ^  ^M^' 
Corol.  2.     Also  take  the  equal  triangles  cet,  can, 
from  the  common  triangle  ced, 

and  there  remains  the  A  ted  =  trapez.  aned. 


THEOREM  XXL 

The  same  being  supposed  as  in  the  last  Proposition  ;  then 
any  Lines  s^,  gq,  drawn  parallel  to  the  two  Tangents » 
shall  also  cot  o^  equal  Spaces. 

That  is, 
the  A  K^G  =  trapez.  anhg. 
the  A  Kqg  =;  trapez.  Ashg, 


0 

For,  draw  the  ordinate  de.    Then 

The  three  sim.  triangles  can,  cde,  cgh,  " 

Vol.  I.  6*  ar« 


b06  CONIC  SECTIONS. 

are  to  each  other  as         ca^  ,  cd^  ,  cg^  ; 
th.  by  dlv.  the  trap,  aned  :  trap,  anhg  :  :  cd^ — ca^  :  cg^ — ca-  , 
But,  by  theor.  1,         de^  :  cq,^  : :  cd^ — ca^  :  cg^ — ca»  ; 

theref.  by  equ.  trap,  aned  :  trap,  anhg  :  :  de*  :  gq,2. 
But,  by  sim.  As,  tri.  ted  :  tri.  kqg  :  :  de^  :  gq*  ; 
theref.  by  equal.        aned  :  ted  :  :       anhg    :  k^g. 

But,  by  cor.  2  theor.  20,  the  trap,     aned  =  A  ted  ; 

and  therefore  the  trap.       anhg  =  A  k^g. 

In  like  manner  the  trap.     an^|:  =  A  Kgg.     q.  e.  d. 

Corol.  1.     The    three    spaces    anhg,  tehg,  kqg  are   all 
equal. 

Corol.  2.  From  the  equals  anhg,  k^g, 
take  the  equals  ANhg^  Kqg, 
and  there  i-emains  ghuG  =  gq<iG, 

Corol.  3.     And  from  the  equals  ghuG,  gq^a, 
take  the  common  space  gqLHGy 
and  there  remains  the  A  i-ftH  =  Lqh, 

Corol.  4.     Again,  from  the  equals  kqg,  tehg, 
take  the  common  space  klhg, 
and  there  remains         telk  =  A  l^h. 


Corol:  5.  And  when,  by 
the  lines  kq,  gh,  moving 
with  a  parallel  motion,  kci 
comes  into  the  position  ir, 
where  cr  is  the  conjugate  to 
CA  :  then 


the  triangle  kqg  becomes  the  triaagle  irc, 
and  the  space  anhg  becomes  the  triangle  anc  ; 
and  therefore  the  A  iRc  =  A  anc  =  A  tec. 

Corol.  6.     Also  when  the  lines  kq,  and  hq,  by  moving  with 
a  parallel  motion,  come  into  the  position  cc,  Me, 
the  triangle  lqh  becomes  the  triangle  ccm, 
and  the  space  telk  becomes  the  triangle  tec  ; 
and  theref.  the  A  ceM  =  A  tec  =  A  anc  =  A  iRc. 


THEOREM  XXIL 

Any  Diameter  bisects  all  its  Double  Ordinates,  or  the  Lines 
drawn  Parallel  to  the  Tangent  at  its  Vertex,  or  to  its  Con*, 
iugate  Diameter. 
^  ^  Tbat 


OF  THE  HYPERBOLA 


That  is,  if  q,q  be  paral- 
lel to  the  taagent  te,  or 
to  ce,  then  shall  lq  =  Lq. 


For,  draw  qh,  qh  perpendicular  to  the  transverse. 

Then  by  cor   3  theor.  21,  the  A  l^h  =  A  i^qh  ; 

but  these  triangles  are  also  equiangular  ; 

conseq.  their  like  sides  are  equal,  or  lq  =  Lq. 

Corol.  1.  Any  diameter  divides  the  hyperbola  into  two 
equal  parts. 

For,  the  ordinates  on  each  side  being  equal  to  each  other, 
and  equal  in  number  ;  all  the  ordinates,. or  the  area,  on  one 
side  of  the  diameter,  is  equal  to  all  the  ordinates,  or  the  area, 
on  the  other  side  of  it. 

Corol.  2.  In  like  manner,  if  the  ordinate  be  produced  to 
the  conjugate  hyperbolas  at  q',  q ,  it  may  be  proved  that 
L€i'  =  L^'.  Or  if  the  tangent  te  be  produced,  then  ev  =  ew. 
Also  the  diameter  gceh  bisects  all  lines  drawn  parallel  to  te 
or  Qy,  and  limited  eithef  by  one  hyperbola,  or  by  its  two  con- 
jugate hyperbolas. 


THEOREM  XXni. 

As  the  Square  of  any  Diameter  : 
Is  to  the  Square  of  its  Conjugate  :  : 
So  is  the  Rectangle  of  any  two  Abscisses  : 
To  the  Square  of  their  Ordinate. 
That  is,  ce2    :  ce^    ;  :  el  .  lg  or  cl^  —  ce^ 
For,    draw   the  tangent 
te,  and  produce  the  ordi- 
nate €tL  to  the  transverse 
at  K       Also  draw  qh,   cm 
perpendicular  to  the  trans- 
verse, and  meeting   eg  in 
H   and    M.     Then   similar 
triangles     being     as     the 
squares  of  their  like  sides, 
it  is, 


by 


60S  CONfC  SEGTIOKS. 

by  sim.  triangles,         A  cet  :  A  clk  :  :  ce^   i  cl^  ; 

or,  by  divJsioa,  A  cet  :  trap,  telk  :  ;  ce^  :  cl^—ce*. 

Again,  by  sim.  tri.     .  A  cea  :   A  lqh   :  :  ce^    :  l^^. 

But,  by  cor.  5tbeor  21,     the  A  ccm  =  A  cet, 

and,  by  cor.  4  tbeor.  21,     the  A  lqH=  trap,  telk  ; 

theref.  by  equality,     ce^   ;  ce*    ::  cl* —ce^  :  l^^, 

or         -         -         -     CE^    :  cc*   :  :  el  .  lg  :  lq,*        Q.  e.  t). 

Corol  1.  The  squares  of  the  ordinates  to  any  diameter, 
are  to  one  another  as  the  rectangles  of  their  respective  ab- 
scisses, or  as  the  difference  of  the  squares  of  the  semi-diame- 
ter and  of  the  distance  between  the  ordinate  and  centre.  For 
they  are  all  in  the  same  ratio  of  cfi*  to  ce^ . 

CoroL  2.  The  above  being  the  same  property  as  that  be- 
longing to  the  two  axes,  all  the  other  properties  before  laid 
down,  for  the  axes,  may  be  understood  of  any  two  conjugate 
diameters  whatever,  using  only  the  oblique  ordinates  of  these 
diameters  instead  of  the  perpendicular  ordinates  of  the  axes  ; 
namely,  all  the  properties  in  theorems  6,  7,  8, 16, 17,  20,  21. 

Corol.  3.  Likewise,  when  the  ordinates  are  continued  to 
the  conjugate  hyperbolas  at  q,',  q\  the  same  properties  still 
ohtain,  substituting  only  the  sum  for  the  difference  of  the 
squares  of  ce  and  cl, 

That  is,  ce2    :  cea    :  :  cl^  +  ce^   :  lq'3  . 

And  80    L^3  :  jt^'s   :  ;  Qj^a  __  q^s  .  cl^  -f-  ces  . 

Corol.  4.  When  by  the  motion  of  lq'  parallel  to  itself,  that 
line  coincides  with  ev,  the  last  corollary  becomes 
ces    :   ce2    : :  gcE^  :  ev^, 
or  cc3    :  Ev2  :  :       1      :     2, 
orce     :  EV     :  :       1     :^2, 
or  as  the  side  of  a  square  to  its  diagonal. 
That  is,  in  all  conjugate  hyperbolas,  and  all  their  diameters, 
any  diameter  is  to  its  parallel  tangent,  in  the  constant  ratio  of 
the  side  of  a  square  to  its  diagonal. 


THEOREM  XXIV. 

If  any  Two  Lines,  that  any  where  intersect  each  other,  meet 
the  Curve  each  in  Two  Points  ;  theft 
The  Rectangle  of  the  Segments  of  the  one  : 
Is  to  the  Rectangle  of  the  Segments  of  the  other  :  : 
As  the  Square  of  the  Diam.  Parallel  to  the  former  : 
To  the  Square  of  the  Diam.  Parallel  to  the  latter. 

TK«t 


OF  THE  HYPERBOLA, 


That  is,  if  cr  and 
Cr  be  parallel  to  any 
two  lines  vhq,  p^q ; 
then  shall  cr^  :  cr*  :  : 
FH  .  Hq  :  pu,  Hq. 


For,  dravv  the  diameter  che,  and  the  tangent  te,  and  its 
parallels  pc,  ri.  mh,  meeting  the  conjugate  of  the  diameter 
CR  in  the  points  t,  k,  i.  m.  Then  because  similar  triangles 
are  as  the  squares  of  their  like  sides,  it  is, 

by  sim.  triangles,     cr*  :  gp*    :  :  A  cri  :  Agpk, 

and         -         -  CR*  :  gh*  :  :  A  cri  :  Aghm  ; 

theref.  by  division,  cil*  :  gp*  — gh*  :  :  cri  :  kphm. 

Again,  by  sim.  tri.     ce*  :  ch*  :  :  Acte  :  Acmh  ; 

and  by  division,        ce*  :  ch*  —  ce*  :  :  Acte  :  tehm. 

But,  by  cor.  6  theor.  21,  the  A  cte  =  A  cir, 

and  by  cor.  1  theor.  21,  tehg  =  kphg,  or  tehm  =  kphm  ; 

theref.  by  equ.  ce*  :  ch*  —  ce*  :  :  cr*  gp*  —  gh*  or  ph  .  hq. 

In  like  manner  ce*  :  ch*  —  ce*  ; :  cr*  :  joh  .  h^. 

Theref.  by  equ.  cr*  :  cr*  :  :  ph  .  na  •*  p^  >  Hy.  q.  e.  d. 

Corol.  1.  In  like  manner,  if  any  other  line  p  h'  q,  parallel 
to  cr  or  to  pq,  meet  phq  ;  since  the  rectangles  ph'q,  p  h  q' 
are  also  in  the  same  ratio  of  cr*  to  cr*  ;  therefore  the  rect. 
PHQ  ;  ptLq  :  :  PH'Ci  :  pviq. 

Also,  if  another  line  p'^a'  be  drawn  parallel  to  pq  or  cr  ; 
because  the  rectangles  p'/iq'  plicf  are  still  in  the  same  ratio, 
therefore,  in  general  the  rectangle  ph^  :  pviq  :  :  p'/ia'  :  ph^ . 
That  is,  the  rectangles  of  the  parts  of  two  parallel  lines,  are 
to  one  another,  as  the  rectangles  of  the  parts  of  two  other 
parallel  lines,  any  where  intersecting  the  former. 

CoTol.  2.  And  when  any  of  the  lines  only  touch  the  curve, 
instead  of  cutting  it,  the  rectangles  of  such  become  squares, 
and  the  general  property  still  attends  them. 

That 


510 


CONIC  SECTIONS. 


That  is, 

cr2  :  cr2  :  :  te^  :  tc^, 

or     CR     :  cr    :  :  TE    :  tCj 

and  CR    :  cr    :  :  tE     :  te. 


Carol.  3.    And  hence  te  :  tc  :  :  iE,  :  te. 


THEOREM  XXV. 


If  a  Line  be  drawn  through  any  Point  of  the  Curves,  Parallel 
to  either  of  the  Axes,  and  terminated  at  the  Asymptotes  ; 
the  Rectangle  of  its  Segments,  measured  from  that  Point, 
will  be  equal  to  the  Square  of  the  Semi-axis  to  which  it  is 
parallel. 


That  is, 
the  rect.  hek  or  hck  =  ca^  , 
aad  rect.  hEk  or  hek  =  ca^  . 


For,  draw  al  parallel  to  ca,  and  ol  to  ca.     Then 
'     by  the  parallels,      ca^  :  ca^  or   al^  :  :  cd^  :  dh^  ; 

and  by  the«>r.  2,       ca^  :  ca^  :  :  cd^  —  ca^  :  de^  ; 

theref.  by  subtr.      ca^  :  ca^  :  :  ca^  :  dh^  —  de^  or  hek. 

But  the  antecedents  ea^  ,   ca^  are  equal, 

theref.  the  consequents      ca^  ^    hek  must  also  be  equal. 
In  like  manner  it  is  again, 

by  the  parallels,  ca^  :  ca^  or  al^  :  :  cd^  :  dh^  ; 

and  by  theor.  3,  ca^  :  ca^  :  :  cd^  -|-  ca^  :  dc  ^  ; 

theref.  by  subtr.  ca^  :  ca^  :  :  ca^  :  dc^  —  dh^  or  hck. 
But  the  antecedents  ca^  ,  cas  are  the  same, 
theref.  the  conseq.    ca^  ,  hck  must  be  equal. 
In  like  manner,  by  changing  the  axes,  is  fiEk  or  hek  =  ca^. 

Corol.  1.  Because  the  rect.  hek  =  the  rect.  hck. 
therefore  eh  :  ch  :  :  ck  :  ek. 
And  consequently  he  :  is  always  greater  than  ue. 
Corol.  2.    The  rectangle  ^  ek  =  the  rect.  uEk. 
For,  by  sim.  tri.  eH  :  eh  i  :  zk  :  ek. 

SCHOLIUM» 


OF  THE  HYPERBOLA.  511 

SCHOLIUM. 

It  is  evident  that  this  proposition  is  general  for  any  Kne  ob- 
lique to  the  axis  also,  namely,  that  the  rectangle  of  the  seg- 
ments of  any  line,  cut  by  the  curve,  and  terminated  by  the 
asymptotes,  is  equal  to  the  square  of  the  s^rui-diameter  to 
which  the  line  is  parallel.  Since  the  demonstration  is  drawn 
from  properties  that  are  common  to  all  diameters. 

THEOREM  XXVI. 

All  the  rectangles  are  equal  which  are  made  of  the  Segments 
of  any  Parallel  Lines  cut  by  the  Curve,  and  hmited  by  the 
Asymptotes. 


That  is, 
therect.  hek  =  hck. 
and  rect.  hsk  =  hek. 


For,  each  of  the  rectangles  hek  or  hck  is  equal  to  the 
square  of  the  parallel  semi-diameter  cs  ;  and  each  of  the  rect- 
angles hKk  or  hek  is  equal  to  the  square  of  the  parallel  semi- 
diameter  ci.  And  therefore  the  rectangles  of  the  segments 
of  all  parallel  lines  are  equal  to  one  another.  Q.  e.  d. 

Corol  1.  The  rectangle  mek  being  constantly  the  same, 
whether  the  point  e  is  taken  on  the  one  side  or  the  other  of 
the  point  of  contact  i  of  the  tangent  parallel  to  hk,  it  follows 
that  the  parts  he,  ke,  of  any  line  hk,  are  equal. 
'  And  because  the  rectangle  hck  is  constant,  whether  the 
point  e  is  taken  in  the  one  or  the  other  of  the  opposite  hyper- 
Solas,  it  follows,  that  the  parts  He,  kc,  are  also  equal. 

GoroL  2.  And  when  hk  comes  into  the  position  of  the  tan-  ■ 
gent  DiL,  the  last  corollary  becomes  il  =  id,  and  im  =  in,  and 

LM  =  DN. 

Hence  also  the  diameter  cir  bisects  all  the  parallels  to  dl, 
which  are  terminated  by  the  asymptote,  namely  rh  =  rk. 

Corol. 


bit 


CONIC  SECTIONS. 


CoroL  3.  From  the  proposition,  and  the  last  corollary,  it 
follows  that  the  constant  rectangle  hek  or  ehe  is  =  il^  .     And 
the  equal  constant  rect.  hck  or  ene  =  mln  or  im^  —  it^. 
CoroL  4.  And  hence  il  =  the  parallel  semi-diameter  C3> 
For,  the  rect.  ehe  =  il^  , 
and  the  equal  rect.  ene  =  im^  —  jl^, 
theref.  il^  =  im^  —  il^  ,  or  im^  =  2il^  ; 
but,  by  cor.  4  theor.  23,       im^  —  2gs2, 
and  therefore        -        -       il    =  cs. 
And  so  the  asymptotes  pass  through  the  opposite  angles  of  all 
>the  ioscribed  parallelograms. 

THEOREM  XXVII. 

The  rectangle  of  any  two  Lines  drawn  from  any  Point  in  the 
Curve,  Parallel  to  two  given  Lines,  and  Limited  by  the 
Asymptotes,  is  a  Constant  Qjaantity. 

That  is,  if  ap,  eg,  di  be  parallels, 
as  also         AQ.,  EK,  DM  parallels, 
then  shall  the  rect.  paq  =  rect.  gek  =  rect.  JP^f . 


For,  produce  ke,  md  to  the  other  asymptote  at  h,  l. 
Then,  by  the  parallels,  he  :  ge  : :  ld  :  id  ; 
but       -         -         -  BK  :  EK  :  :  DM  :  DM  ; 

theref.  the  rectangle       hek  :  gek  :  :  ldm  :  idm. 
But,  by  the  last  theor.  the  rect.  hek  =  ldm  ; 
and  therefore  the  rect.  gek  =  idm  =  paq. 

THEOREM  XXVm. 


Every  Inscribed  Triangle,  formed  by  any  Tangent  and  the 
two  Intercepted  Parts  of  the  Asymptotes,  is  equal  to  a 
Constant  (Quantity  ;  namely  Double  the  Inscribed  Paral- 
lelogram. 

That 


OP  THE  HYPERBOLA. 

2  paral.  ox. 


518 


That  is,  the  triangle  cts 
For,  since  the  tan«fent  ts  is 
bisected  by  the  point  of  contact 
E,  and  EK.is  parallel  to  to,  and 
GE  to  cK  ;  therefore  ck,  ks,  ge 
are  all  equal,  as  are  also  cg,  gt, 

KE.      Consequently  the  triangle       .__ 

GTE    =   the   triangle    kes,  and         '  ^^  S 

each  equal  to  half  the  constant  inscribed  parallelogram  ck. 
And  therefore  the  whole  triangle  cts,  which  is  composed  of 
the  two  smaller  triangles  and  the  parallelogram,  is  equal  to 
double  the  constant  inscribed  parallelogram  gk.  q.  e.  d. 


THEOREM  XXIX. 

If  from  the  Point  of  ' 

Intersections  of  the  Curve  witii  a  Line  parallel  to  the 
Tangent,  three  parallel  Lines  be  drawn  in  any  Direction, 
and  terminated  by  either  Asymptote  ;  those  three  Lines 
shall  be  in  continued  Proportion. 

That  is,  if  hkm  and  the 
tangent  il  be  parallel,  then 
are  the  parallels  dh,  ei,  gk 
in  continued  proportion 


For,  by  the  parallels,  ei  :  il  :  :  dh  :  H»f  ; 

and,  by  the  same  ei  :  il  :  :  gk  :  km  ; 

theref  by  compos,     ei^  :  il^  :  :  dh  .  gk  :  hmk  ; 
but,  by  theor.  26,  the  rect.  hmk  =  il^  ; 
and  theref  the  rect.        dh     gk  =  ei^, 
©r  -         -         -         dh  :  ei  :  :  ei  :  gk.  q. 


THEOREM  XXX. 

Draw  the  semi-diameters  ch,  cin,  ck  *, 
Then  shall  the  sector  ©hi  =  the  sector  cik. 


Vol.  I. 


For, 


1^14  CONIC  SECTIONS. 

For,  becaujse  hk  and  all  its  parallels  are  bisected  by  cm, 

therefore  the  triangle     cnh  =  tri.  cnk, 

and  the  seju^ment  inh  =  seg    ink  ; 

consequently  the  sector  cih  =  sec.  cik. 
CoroL     If  the   geometricals  dh,  ei,   gk  be  parallel  to  the 
©ther  asymptote,  the   spnces   i>hie,   eikg   will  be  equal  ;  for 
they  are  equal  to  the  equal  sectors  chi,  cik. 

So  that  by  taking  any  geonietricals  cd,  ce,  cg,  &c.  and 
drawing  dh,  ei,  ok,  kc.  parallel  to  the  other  asymptote,  a^ 
also  the  radii  ch,  ci,  ck  ; 

then  the  sectors  chi,  cik,  &c. 

or  the  spaces  DHiE,  eikg,  &,c. 

will  be  all  equal  among  themselves. 

Or  the  sectors  chi,  chk,  &c. 

or  the  sv>aces  dhie,  dhkg,  &c. 

will  be  in  arithmetical  progression. 
And  therefore  these  sectors,  or  Rpaces,»will  be  analogous  to 
the  logarithn^s   of  the  lines  or  bases  cd,  ce,  cg,  kc.  ;  namely 
CHI  or  DHIE  the  log.  of  the  ratio  of 
GD  to  GE,  or  of  CE  to  CG,  &c.  ;  or  of  EI  to  DH,  or  of  gk  to  Ei„&c. ; 

and  ciiK  or  dhkg  the  log.  of  the  ratio  of 

CD  to  CG,  &C.  or  of  GK  tO  DH,  &C. 


OF  THE  PARABOLA. 

THEOREM  I. 

The  Abscisses  are   Proportional     to    the    Squares  of  their 
Ordinates. 

Let  avm  be  a  section  through 
the  axis  of  the  cone,  and  agih  a 
parabolic  section  by  a  plane  per- 
pendicular to  the  former,  and 
parallel  to  the  side  vm  of  the 
cone  ;  also  let  afh  be  the  common 
intersection  of  the  two  planes,  or 
the  axis  of  the  parabola,  and  fg, 
hi  ordinates  perpendicular  to  it. 

Then  it  will  be,  as  af  :  ah  :  :  fg^  :  hi^. 

t'or,  through  the  ordinates    fg,  hi  draw  the  circular  sec- 
tions, KGL,  Miw,  parallel  to  the  base  of  the  cone,  having  kl, 


OF  THE  PARABOLA. 


£16 


MN  for  their  diameters,  to  which  fg,  hi  are  ordinates,  as 
well  as  to  the  axis  of  the  parabola.  *c 

Then,  hy  similar  triangles,  af  :  ah  :  :  fl  :  hw  ; 
but,  because  of  the  parallels,  kf  =  mh  ; 

therefore         -         -         -       af  :  ah  :  :  kf  .  fl  :  mh  .  hw. 
But,  by  the  circle,  kf  .  fl  =  fg",  and  mh  .  hn  =  hi^  ; 
Therefore       -         -         -       af  :  ah  :  :  fg     :  hi^.        q.  e.  o. 

fg3        hi* 

Corol.  Hence  the  third  proportional or is  a  constant 

af  ah 

quantity,  and  is  equal  to  the  parameter  of  the  axis  by  defin, 
16. 

Or  AF  :  fg  :  :  FG  :  p  the  parameter 
Or  the  rectangle  p  .  af  =  fq^. 


THEOREM  U. 


As  the  Parameter  of  the  Axis  : 
Is  to  the  Sum  of  any  Two  Ordinates  :  : 
So  is  the  Difffirence  of  those  Ordmates  : 
To  the  Difference  of  their  Abscisses  : 


J 

^--n 

L 

'^ 

/     ^ 

N 

d 

] 

t      V 

That  is, 
F  :  GH  -|-  DE  :  :  gh  —  de  ; 
Or,  p  :  Ki  :  :  ih  :  is. 


For,  by  cor.  theor,  1,  p  .  ag  =  gh^, 
and         -         -         -        p  .  AD  =  de2 
theref  by  subtraction,     p  .  dg  =  gh"  —  de^ 
Or,         -         -         -        p  .  dg  =  KI .  IH, 
therefore         -         -        p  :  ki  :  :  ih  :  dg  or  ef.  <i.  e.  b. 

Corol   Hence  because  p  .  ei  =  ki  .  ih, 
and,  by  cor.  theor.  1,      p  .  ag  =  gh^  , 
therefore         -         -        ag  :  ei:  :  gh^  :  ki  .  ih. 

So  that  any  diameter  ei  .is  as  the  rectangle  of  the. seg- 
ments ki,  ih  of  the  double  ordinate  kh. 

THEOREM  nr. 


The  Distance  from  the  Vertex  to  the  Focus  is  equal  to  \  of 
Hie  Parameter,  or  to  Half  th«  Ordinate  at  the  Focus. 

That 


316 


CONIC  SECTIONS. 


That  is, 

where  f  is  the  focus. 

/  E 

For,  the  general  property  is  af  :  fe  :  :  fe  :  p. 

But  by  dt^fiuitioD  17,       -        ef  =  ^p  ; 
therefore  also        .    -         -        af  =  ^fe  =  ip, 


THEOREM  IV. 


A  Line  drawn  from  the  Focus  to  any  Point  in  the  Curve,  is 
e(}ual  to  the  Sum  of  the  Focal  Distance  and  the  Absciss  of 
the  Ordinate  to  that  Point. 

G 

That  is,  ^--^■" 

fe  =  FA  -\-  AD  =  GD,  y^        ^~ 

taking  ag  =  af.  /  "E  < 


For.  since  fd  —  ad  co  af, 
theref  by  squaring,     fd^  =  af^  —  2af  .  ad  +  A»^, 
But,  by  cor.  theor.  1,  de^  =  p  .  ad  =  4af  .  ad  ;  j  \ 

theref  by  addition,      fd^  -|-  de=  =  af^  -f-  2af  .  ad  4"  At)3, 
But,  by  right- ang.  tri.  fd^  -{-  de^  =  fe^  ; 
therefore         -  -  fe^  =  af^  -{- 2af  .  ad  +  ad^, 

and  the  root  or  side  is  fe    —  af  +  ad, 
or  -         -         -  i,E  =  GD,  by  taking  ag  =  af. 

^.   E.   ©. 

Corol  1.  If,  through  the  point  G,  the  IIHH  G-  HHH 
line  GH  be  drawn  perpendicular  to  the 
axis,  it  is  called  the  directrix  of  the  pa- 
rabola. The  property  of  which,  from 
this  theorem,  it  appears,  is  this  :  That 
driiwing  any  line  he  parallel  to  the  axis, 
HK  is  always  equal  to  fe  the  distance  of  J^  jq-       ^ 

the  focus  from  the  point  e, 

'Corol.  2.  Hence  also  the  curve  is  easily  described  by  points. 
Namely  in  the  axis  produced  take  ag  =  af  the  focal  dis- 
tance, and  draw  a  number  of  lines  ee  perpendicular  to  the 
axis  AD  ;  then  with  the  distances  gd,  gd,  gd,  &c.  as  radii 
and  the  centre  f,  draw  arcs  crossing  the  parallel  ordinates 
in  F,  E,  E,  &G.     Then  draw  the  curve  through  all  the  points, 

£;    £;    £. 

THEOREM 


y 

\ 

■ 

/E 

F    E\ 

/ 

/B 

D       E 

\ 

OF  THE  PARABOLA. 


517 


THEORKM.V- 

If  a  Tangent  be  drawn  to  any  Point  of  the  Parabola,  meet- 
ing the  Axis  produced  ;  and  if  an  Ordinate  to  the  Axis 
be  drawn  from  the  Point  of  Contact  ;  then  the  Ahscigs  of 
that  Ordinate  will  be  equal  to  the  External  Fart  of  the 
Axis, 


That  is, 
if  Tc  touch  the  curve 
at  the  point  c  ; 
then  is  at  =  am. 


For,  from  the  point  t,  draw  any  line  cutting  the  curve  in  the 
the  two  points  e,  h  :  to  which  draw  the  ordinates  de,  gh  ;  also 
draw  the  ordinate  mc  to  the  point  of  contact  c. 
AD  :  AG  :  :  de2   :  gh^  ; 
:  tg3  ;  :  de^  :  gh^  ; 
;  AG  :  :  TD^  :  tg2  ; 

pg  :  :  TD^  :  tg^  — td^  or  dg.(td-|-tg), 
:  TD  :  :  TD  :  td-|-tg  ; 
:  AT  :  :  td'  :  tg, 
AT  :  :  AT  :  AG  ; 


Then,  by  th 

and  by  sim   tri.  td- 
theref.  by  equality, ad 

and,  by  division,  ad 

or          -         -  ad 

and,  by  division,  ad 

and  again  by  div.  ad 


AT  is  a  mean  propor.  between  ad,  ag. 

Now  if  the  line  th  be  supposed  to  revolve  about  the 
point  T  ;  then,  as  it  recedes  farther  from  the  axis,  the  points  e 
and  H  approach  towards  each  other,  the  point  e  descending  and 
the  point  H  ascending,  till  at  last  they  meet  in  the  point  c, 
when  the  line  becomes  a  tangent  to  the  curve  at  c  And 
then  the  points  d  and  g  meet  in  the  point  m,  and  the  ordinates 
DE,  gh  in  the  ordinates  cm.  Consequently  ad,  ag,  becoming 
earh  equal  to  am,  their  mean  proportional  at  will  be  equ^J  to 
the  absciss  am.  That  is  the  external  part  of  the  axis,  cut  off 
by  a  tangent,  is  equal  to  the  absciss  of  the  ordinate  to  the  point 
©f  contact.  Q.  E.  D. 

THEOREM  VI. 


If  a  Tangent  to  the  Curve  meet  the  Axis  produced  ;  then 
the  Line  drawn  frbm  the  Focus  to  the  i  oint  of  Contact,  will 
be  equal  to  the  Distance  of  the  Focus  from  the  Intersection 
of  the  Tangent  and  Axis. 

,        That 


&u 


aONIC  SECTIONS. 


That  is, 

FG  =  FT. 


For,  draw  the  ordinate  dc  to  the  point  of  contact  c. 

Then,  by  theor.  5,  at  =  ad  ; 
therefore  -         ft  =  af  +  ad. 

But,  by  theor.  4,         fc  =  af  -|-  ad  ; 
ther^f.  by  equaUty,      fc  =  ft. 

Corol.    1.      If  CG  be  drawn  perpendicular  to  the  curve,  or  to 
the  tangent,  at  c  ;  then  shall  fg  =  fc  =  ft. 

For,  draw  fh  perpendicular  to  tc,  which  will  alsjo  bisect 
tc,  because  ft  =  fc  ;  and  therefore,  by  the  nature  of  the 
parallels,  fh  also  bisects  tg  in  f.     And  consequeaily    tc  = 

FT  =   FC. 

So  that  F  is  the  centre  of  a  circle  passing  through  r,  c,  g. 

Corol.  2.     The  tangent  at  the  vertex  ah  is  a  mean  propor- 
tional between  af  and  ad. 

For,  because  fht  is  a  right  angle, 
therefore      -     ah  is  a  mean  between  af,  at, 
or  between  -     af,  ad,  because  ad  =  at. 
Likewise,     -     fij  is  a  mean  between  fa,  ft, 
or  between  fa,  fc 


Corol.  3. 


The  tangent  tc  makes  equal  angles  with  fc  and 
the  axis  ft. 


For,  because  ft  =  fc,  ^ 

Therefore  the  ^  fit  =  ^ftc. 

Also,  the  angle  gcf  =  the  angle  gck, 

drawing  ick  parallel  to  the  axis  ag. 
Corol.  4.  And  because  the  angle  of  incidence  gck  is  = 
the  angle  of  reflection  gc  f  ;  therefore  a  ray  of  light  falling 
on  the  curve  in  the  direction  kc,  will  be  reflected  to  the  fo- 
cus F.  That  is,  all  rays  parallel  to  the  axis,  are  reflected  to 
the  focus,  or  burning  point. 

THEOREM 


OP  THE  PARABOLA. 


51« 


THEOREM  VU. 


if  there  be  any  Tangent,  and  a  DouMe  Ordinate  drawn  from 
the  I'oint  of  Contact,  and  also  any  Line  parallel  to  the  Axis, 
limited  hy  the  Tangent  and  Double  Ordinate  :  then  shall 
the  Curve  divide  that  Line  in  the  same  Ratio,  as  the  Line 
divides  the  double  Ordinate. 


That  is, 
;  EK  :  :  CK 


For.  by  sim.  triangles, 
but  by  the  def  the  param 
therefore,  by  equality, 
But,  by  theor.  2, 
therefore,  by  equality, 
and  by  division. 


3S,     CK  :  KI  :  :  cd 

DT  or  2da  ; 

•am.     p  :  CL  :  :  CD 

:  2da; 

p  :  CK  : :  cl 

:  KI. 

-    p  :  CK  :  :  KL 

.  ke; 

CL  :  KL  :  :  ki 

ke; 

CK  :  KL  :  :  ie 

:  EK.              d.   E. 

THEOREM  VIIL 

Th6  same  being  supposed  as  in  theor.  7  ;  then  shall  the  Ex- 
ternal Fart  of  the  Line  between  the  Curve  and  Tangent 
be  proportional  to  the  Square  of  the  intercepted  Part  of 
the  Tangent,  or  to  the  bquare  of  the  intercepted  Fart  of 
the  Double  Ordinate. 


That  is,  IE  is  as  ci^  or  as  ck^ 
and  ic,  TA,  ON,  pl,  &c. 
are  as  ci^,  ct^,  co^,  cp,  ^:c. 
•r  as  gk2,  cd2,  cm^,  cl^,  &e. 


For,  by  theor.  7 
or,  by  equality. 
But,  by  cor.  th.  2, 
therefore 


IE  :  EK  : 

IE    :   EK 


EK  is  as  the  rect.  ck  .  kl, 
IE  is  as  ck3,  or  as  ci2.  q.  k.  d- 

Corol.  As  this  property  is  common  to  every  position  of  the 
tangent,  if  the  lines  ie,  ta,  on.  &c.  be  appended  on  the  points 
I,  T,  o,  ^c.  and  moveable  about  them,  and  of  surh  lengths  as 
that  th«ir  extromities  e,  a,  n,  kc.  be  in  the  curve  of  a  para- 
bola 


529  CONIC  SECTIONS. 

.  bola  in  some  one  position  of  the  tangent  ;  then  making  the  taa* 
gent  revolve  about  the  point  c,  it  appears  thnt  the  extremities 
E,  A,  N,  ^c.  will  always  fortn  the  curve  of  some  parabola,  in 
every  position  of  the  tangent. 

THEOREM  IX. 

The  Abscisses  of  any  Diamr  tf^r,  are  as  the  Squares  of  their 
Ordinates. 


That  is,  CQ,  or,  cs,  &c. 
are  as  q,E^ ,  ra^  ,  sn^  ,  lk,c. 
Op       CQ  :  CR  :  :  ^e^  :  ra, 
&c. 


For,  draw  the  tangent  ct,  and  the  externals  ei,  at,  no,  &c, 
parallel  to  the  axis,  or  to  the  diameter  cs. 

Then,  because  the  ordinates  qE,  ra,  sn,  &c.  are  parallel  to 
the  tangent  ct,  by  the  definition  of  them,  therefore  all  the  fi- 
gures iQ,  TR,  OS,  &c.  are  parallelograms,  whose  opposite  sides 
are  equal  ; 

namely,         -         -         ie,    ta,    on,    &c. 
are  equal  to  -         cq,  gr,    cs,    &c. 

Therefore,  by  theor.  8,  cq,  cr,    cs,    &c. 
areas  -         -         ci^,  ct^,  co^,  &c. 

or  as  their  equals      :     qe^,  ra^,  sn^,  &c.  q.  e.  ». 

CoroL  Here  like  as  in  theor.  2  the  difference  of  the  abscis- 
ses is  as  the  difference  of  the  squares  of  their  ordinates.  or  as 
the  rectangles  under  the  sum  and  difference  of  the  ordinates, 
the  rectangle  of  the  sum  and  difference  of  the  ordinates  being 
equal  to  the  rectangle  under  the  difference  of  the  abs(  i^ses 
and  the  parameter  of  that  diameter,  or  a  third  proportional  t© 
any  absciss  and  its  ordinate. 


THEOBExM 


OF  THE  PARABOLA. 

THEOREM  X. 


621 


tf  a  Line  be  drawn  parallel  to  any  Tangent,  and  cut  the  Curve 
in  two  Points  ;  then  it  two  Ordinates  be  drawn  to  the  In- 
tersections, and  a  third  to  the  Point  of  Contact,  these  three 
Ordinates  will  be  in  Arithmetical  Progression,  or  the  Sum 
of  the  Extremes  will  be  equal  to  Double  the  Mean. 


That  is, 

BG  +  HI  =  2CD. 


.^ 

A 

% — 1 

/ 

H              1 

L    ] 

L                        lu 

For,  draw  ek  parallel  to  the  axis,  and  produce  hi  to  l. 
Then,  by  sim.  triangles,  ek  :  hk  :  :  td  or  2ad  :  cd  ; 
but,  by  theor.  2,     -         ek  :  hk  :  :  kl  :  p  the  param. 
theref.  by  equality,       2ad  :  kl  :  :  cd  :  p. 
But,  by  the  defin.  2ad  :  2cd  :  :  cd   :  p  ; 

theref.  the  2d  terms  are  equal,  kl  =  2cd, 
that  is,  -  -         EG  -|-  HI  =  2cD.  q. 


E.  o. 


CoroU  When  the  p©int  e  is  on  the  other  side  of  ai  ;  then 

HI  — GE  =  2CD. 

THEOREM  XI. 

Any  Diameter  bisects  all  its  Double  Ordinates, or  Lines  parallel 
io  the  tangent  at  its  Vertex. 


That  is, 

ME  =  MH. 


For,  to  the  axis  ai  draw  the  ordinates  eg,  cd,  hi,  and  mn 
parallel  to  them,  which  is  equal  to  cd. 
V9&.  I.  67  Then 


6n 


CONIC  SECTIONS. 


Then,  by  theor.  10,  2mn  or  2cd  =  eg  +  hi> 
therefore  m  is  the  middle  of  eh. 

And,  for  the  same  reason,  all  its  parallels  are  bisected. 

Q.    E.   D. 

ScHOL.  Hence,  as  the  abscisses  of  any  diameter  and  their 
ordinates  have  the  same  relations  as  those  of  the  axis,  namely, 
that  the  ordinates  are  bisected  by  the  diameter,  and  their 
squares  proportional  to  the  abscisses  ;  so  all  the  other  pro- 
perties of  the  axis  and  its  ordinates  and  abscisses,  before  de- 
monstrated, will  likewise  hold  good  for  any  diameter  and  its 
ordinates  and  abscisses.  And  also  those  of  the  parameters, 
understanding  the  parameter  of  any  diameter,  as  a  third 
proportional  to  any  absciss  and  its  ordinate.  Some  of  the 
most  material  of  which  are  demonstrated  in  the  following 
theorems. 

THEOREM  Xn. 

The  Parameter  of  any  Diameter  is  equal  to  four  Times  the 
Line  drawn  from  the  Focus  to  the  Vertex  of  that  Dia- 
meter. 


That  is,  4fc  =  p, 
the  param.  of  the  diam.  cm. 


For,  draw  the  ordinate  ma  parallel  to  the  tangent  ct  ;  also 
CD,  MN  perpendicular  to  the  axis  an,  and  fh  perpendicular  to 
the  tangent  ct. 

Then  the  abscisses  ad,  cm  or  at,  being  equal,  by  theor.  6, 
the  parameters  will  be  as  the  squares  of  the  ordinates  cd,  ma 
or  CT,  by  the  definition  ; 

that  is,         -         -         p  :  p  : 

But,  by  sim.  tri.  -      fh  :  ft  : 

therefore     -         -         p  :  p  : 

But,  by  cor.  2,  th.  6,  fh^  =fa  . 

therefore     -         -        p  :  p  : 

or,  by  equality     -         p  :  p  : 

But, 


CD^ 

:  CT^. 

:  CD 

CT  ; 

FH3 

:  ft2. 

ft  ; 

FA  . 

FT  :  ft 

FA  : 

ft  orFc 

But,  by  theor.  3, 
and  therefore     - 


op  THE  PARABOLA. 

p  =  4fa, 

p  =  4ft  or  4fc. 


523 


q,.  E.  D. 


Corel.  Hence  the  parameter  p  of  the  diameter  cm  is  equal 
to  4fa  -f  4ad,  or  to  p  -i-  4ad,  that  is,  the  parameter  of  the 
axis  added  to  4 Az>. 


THEOREM  Xni. 


If  an  Ordinate  to  any  Diameter,  pass  through  the  Focus,  it 
will  be  equal  to  Half  its  Parameter  ;  and  its  Absciss  equal 
to  One  Fourth  of  the  same  Parameter. 


That  is,  CM  =  ^p, 

and  ME  =  |p« 


For,  join  fc,  and  draw  the  tangent  ct. 

By  the  parallels,  cm  =.  ft  ; 

and,  by  theor.  6,  fc  =  ft  ; 

also,  by  theor.  12,  fc  =  Jp  ; 

therefore      -      -  cm  =  ip,  . 

Again,  by  the  defin.  cm  or  |p  :  me  :  :  me  :  p, 

and  consequently       me  =  ^p  =  2cm.  ft*  e.  d. 

CoroL  1.  Hence,  of  any  diameter,  the  double  ordinate 
which  passes  through  the  focus,  is  equal  to  the  parameter,  or 
to  quadruple  its  absciss. 


CoroL  2.  Hence,  and  from  cor.  1 
to  theor.  4,  and  theor.  6  and  12,  it 
appears,  that  if  the  directrix  gh  be 
drawn,  and  any  lines  he,  he,  pa- 
rallel to  the  axis  ;  then  e?ery  parallel 
he  will  be  equal  to  ef,  or  ^  of  the 
parameter  of  the  diameter  to  the 
point  e. 


THEOREM 


524 


CONIC  SECTIONS. 


THEOREM  XIV. 

If  there  be  a  Tangent,  and  any  Line  drawn  from  the  Point 
of  Contact  and  meetinfic  the  Curve  in  some  other  Point,  as 
also  another  Line  parallel  to  the  Axis,  and  limited  by  the 
First  Line  and  the  Tangent  :  then  shall  the  Curve  divide 
this  Second  Line  in  the  same  Ratio,  as  the  Second  Line 
divides  the  first  Line. 


That  is, 
IE  :  EK  :  :  ck  :  kl. 


For,  draw  lp  parallel  to  ij£,  or  to  the  axis. 
Then  by  theor.  8,         ie  :  pl  :  :  ci2  :  cp^, 
or,  by  sim.  tri.     -         ie  :  pl  :  :  ck^  :  cl^  . 
Also,  by  sim.  tri.  ik  :  pl  :  :  ck    :  cl, 

or     -         -         -  IK  :  PL  :  :  ck^  :  ck  .  cl  ; 

therefore  by  equality,  ie  :  ik  :  :  ck  :  cl  :  cl^  ; 
or,     -         -         -         IE  :  IK   :  :  CK  :  CL  ; 
and,  by  division,  ie  :  ek  :  :  ck  :  kl. 

CoroL     When  ck  =  kl,  then  ie  =  ek  =  ^ik. 


Q.  E.  D. 


THEOREM  XV. 

If  from  any  Point  of  the  Curve  there  be  drawn  a  Tangent, 
and  also  Two  Right  Lines  to  cut  the  Curve  ;  and  Dia- 
meters be  drawn  through  the  Points  of  Intersection  e  and 
L,  meeting  those  Two  Right  Lines  in  two  other  Points  g,. 
and  K  ;  Then  will  the  Line  kg  joining  these  last  Two 
Points  be  parallel  to  the  Tangent. 


For, 


OF  THE  PARABOLA.  625 

Fbm,  by  theor.  14,  ck  :  kl  :  :  ei   :  ek  ; 

and  by  composition,  ck  :  cl  :  :  ei    :  ki  ; 

and  by  the  parallels  ck  :  cl  :  :  gh  :  lh  ; 

But,  by  sim.  tri.     -  ck  :  cl  :  :  Jti  :  lh  ; 

theref.  by  equal.     -  ki   :  lh  :  :  gh  :  lh  ; 

cont^equently          -  ki  «=  gh, 

and  therefore         -  kg  is  parallel  and  equal  to  ih. 

THEOREM  XVI. 

If  an  Ordinate  be  drawn  to  the  Point  of  Contact  of  any 
Tangent,  and  another  Ordinate  produced  to  cut  the  Tan- 
gent ;  It  will  be,  as  the  Difference  of  the  Ordinates : 

Is  to  the  Difference  added  to  the  external  Part  :  : 

So  is  Double  the  first  Ordinate  : 

To  the  Sum  of  the  Ordinates. 

That  is,  KH  :  KI  : :  KL  :  kg. 
T 


IH     K.  Ja     & 


For,  by  cor.  1  theor.  1,  p  :    dc  : :  dc  :  da. 

and  -         -         -       p  :  2dc  : :  do  :  dt  or  2da. 

But,  by  sim.  triangles,  ki  :    kc  : :  dc  :  dt  ; 

therefore  by  equality,     p  :  2dc  :  ;  ki  :  kc, 

or,  -  -  -        p  :    KI   :  :  KL  :  kc. 

Again,  by  theor.  2,  p  :   kh  :  :  kg  :  kc  ; 

therefore  by  equality,  kh  :    ki  :  :  kl  :  kg.  q.  e.  d. 

Coroh  1.    Hence,  by  composition  and  division, 
it  is,  kh  :  KI    :  :  GK  :  gi, 
and   hi  :  hk  :  :  hK  :  kl, 
also  IH  :  IK    :  :  IK  :  ig  ; 
that  is,  IK  is  a  mean  proportional  between  ig  and  ih. 

€orol.  2.  And  from  this  last  property  a  tangent  can  easily 
be  drawn  to  the  curve  from  any  given  point  i.  Namely,  draw 
iHG  perpendicular  to  the  axis,  and  take  ik  a  mean  proportion- 
al between  ih,  ig  ;  then  draw  kc  parallel  to  the  axis,  and  c 
will  be  the  point  of  contact,  through  which  and  the  given 
point  I  the  tangent  le  is  to  be  drawn. 

THEOREM 


526 


CONIC  SECTIONS. 


THEOREM  XVn. 

If  a  Tangent  cut  any  Diameter  produced,  and  if  an  Ordinate 

•  to  that   Diameter  be  drawn  from  the   Point  of  Contact ; 

then  the  Distance  in  the  Diameter  produced,  between  the 

Vertex  and  the  Intersection  of  the  Tangent,  will  be  equal 

to  the  Absciss  of  that  Ordinate. 

For,  by  the  last  th.  ie  ;  ek  : :  ck  :  kl. 
But,  by  theor.  1 1 ,  CK  =  KL,  ' 

and  therefore        ie  =  ek. 

Corol.  1.  The  two  tangents  ci,  li,  at  the  extremities  of 
any  double  ordinate  cl,  meet  in  the  same  point  of  the  diame- 
ter of  that  double  ordinate  produced.  And  the  diameter 
drawn  through  the  intersection  of  two  tangents,  bisects  the 
line  connecting  the  points  of  contact. 

Corol,  2.  Hence  we  have  another  method  of  drawing  a 
tangent  from  any  given  point  i  without  the  curve,  Namely, 
from  I  draw  the  diameter  ik,  in  which  take  ek  =  ei,  and 
through  K  draw  cl  parallel  to  the  tangent  at  e  ;  then  c  and  l 
are  the  points  to  which  the  tangents  must  be  drawn  from  i. 

THEOREM  XVIII. 

If  a  Line  be  drawn  from  the  Vertex  of  any  Diameter,  to  cut 
the  Curve  in  some  other  Point,  and  an  Ordinate  of  that 
Diameter  be  drawn  to  that  Point,  as  also  another  Ordinate 
any  where  cutting  the  Line,  both  produced  if  necessary  : 
The  Three  will  be  continual  Proportionals,  namely,  the  two 
Ordinates  and  the  Part  of  the  Latter  hmited  by  the  said  Line 
drawn  from  the  Vertex. 

Ml 


That  is,  DE,  GH,  Gi  are 

/ 

'^p.-.TX 

i=^^r 

continual  proportionals^  or 

/^ 

'"^^^];^m 

DE  :  GH  :  :  GH  :  gi.                         / 

Ti"-^ 

O         \^ 

For,  by  theor.  9,     -     -     -     de^ 

:  gh2 

:  :  AD 

:  AG  ; 

and,  by  sim.  tri.       -     -     -     de 

:  GI 

:  :  AD 

:  AG; 

theref.  by  equality,       -     -     de 

;  GI 

:  :  de3 

:   GH3  ; 

that  is,  of  the  three  de,  gh,  gi,  1st  : 

:  3d 

:  :   lst2 

:  U^; 

therefore        »        -        -        ist 

:2d 

::  2d 

:  3d, 

that  is,  -         -        -        -       DE  : 

:  GH 

:  :  GH    ; 

:    GI.     Q.  E^  d. 

Corol, 

OF  THE  PARABOLA. 


527 


Corol,  1.  Or  tbeir  equals,  gk,  gh,  gi,  are  proportionals  ; 
where  ek  is  parallel  to  the  diameter  ad.    . 

CoroL  2.     Hence  it  is  db  :  ag  : :  p    :  gi,  where  p  is 
the  paranaeter,  or      ag  :  gi  : :  de  :  p. 
For,  by  the  defin.      ag  :  gh  :  :  gh  :  p. 
Corol.  3.     Hence  also  the  three  mn ,  aii,  mo,  are  proportion- 
als, where  mo  is  parallel  to  the  diameter,  and  am  parallel  to 
the  ordinates. 

For,  by  theor.  9,   -     mn,  mi,  mo, 
or  their  equals       -     ap,  ag,  ad, 
are  as  the  squares  of  pn,  gh,  de, 
or  of  their  equals        gi,  gh,  gk, 
which  are  proportionals  by  cor.  1 . 

THEOREM  XIX. 

If  a  Diameter  cut  any  Parallel  Lines  terminated  by  the  Curve  : 
the  Segments  of  the  diameter  will  be  as  the  Rectangle  of  the 
Segments  of  those  Lines. 

That  is,  EK  :  EM  :  :  GK .  KL  :  NM  .  MO. 

Or,  EK  is  as  the  rectangle  ck  .  kl. 

For,  draw  the  diameter 
PS  to  which  the  parallels 
CL,  NO  are  ordinates,  and  ' 
the  ordinate  e^  parallel  to 
them, 

Then  ck   is  the  differ- 
ence, and  KL  the  sum  of 
the   ordinates  eq,  cr  ;  also 
NM  the  difference,  and  mo  the  sum  of  the  ordinates  eq,  ns. 
And  the  differences  of  the  abscisses,  are  ^r,  %s,  or  ek,  em. 

Then  by  cor.  theor.  9,  ^r  :  ^s  :  :  ck  .  kl  :  nm  .  mo, 
that  is         -         -  ek  :  em  :  :  CK  .  KL  ;  NM  .  mo. 

Corol.  1.  The  rect.  ck  .  kl  =  rect.  EKand  the  param.  of  ps. 
For  the  rect.   ck  .  kl  =  rect  q.r  and  the  param.  of  ps. 

Corol.  2.  If  any  line  cl  be  cut  by  two  diameters,  ke,  gh  ; 
the  rectangles  of  the  parts  of  the  line,  are  as  the  segments  of 
the  diameters. 

For  EK  is  as  the  rectangle  ck  .  kl. 
and  GH  is  as  the  rectangle  ch  .  hl  ; 
therefore  ek  :  gh  :  :  ck  .  kl  :  ch  .  hl. 

CoroL 


528 


CONIC  SECTIONS. 


Corol.  3.  If  two  parallels,  cl,  no,  be  cot  by  two  diame- 
ters, EM,  Gi ;  the  rectangles  of  the  parts  of  the  parallels,  will 
be  as  the  segments  of  the  respective  diameters. 

For     -         -    '     -     EK  :  EM  :  :  CK  .  KL  :  NM  .  mo, 
and     -         -         -     EK  :  GH  :  :  CK  .  KL  :  CH  .  hl, 
theref.  by  equal.        em  :  gh  :  :  nm  .  mo  :  ch  .  hl. 
Corol.  4.     When  the  parallels  come  into  the  position  of  the 
tangent  at  p,  their  two  extremities,  or  points  in  the  curve  unite 
in  the  point  of  contact  p  ;  and  the  rectangle  of  the  parts  be- 
comes the  square  of  the  tangent,  and  the  same  properties  still 
follow  them. 

So  that,  Ev  :  pv  :  :  fv  :  p  the  param. 
gw  :  pw  :  :  pw  :  p, 
EV  :  gw  :  :  pv^  :  pw^, 
EV  :  GH  :  :  pv^  :  ch  .  HLi 

THEOREM  XX. 

If  two  Parallels  intersect  any  other  two  Parallels  ;  the  Rect- 
angles of  the  Segments  will  be  respectively  Proportional. 
That  i&,  CK  .  KL  :  DK  .  ke  :  :  gi  .  ih  :  ni  .  lo. 


For,  by  cor.  3  theor.  23,  pk  :  qi  :  :  ck  .  kl  :  gi  .  ih  ; 

and  by  the  same,  pk  :  qi  :  :  dk  :  ke  .  ni  .  lo  ; 

theref.  by  equal,  ck  .  kl  :  dk  :  ke  :  :  gi  .  ih  :  ni  .  lo. 

Corol.  When  one  of  the  pairs  of  intersecting  lines  comes 
into  the  position  of  their  parallel  ttingents,  n»eeting  and  limit- 
ing each  other,  the  rectangles  of  their  segments  become  the 
squares  of  their  respective  tangents.  So  that  the  constant 
ratio  of  the  rectangles,  is  that  of  the  square  of  their  parallel 
tangents,  namely, 
CK  .  KL  :  pK  .  KE  :  :  tan^  .  parallel  to  cl  :  tang^  .  parallel  to  de, 

THEOREM  XXI. 


If  there  be  Three  Tangents  intersecting  6ach  other ; 
Segments  will  be  in  the  same  Proportion. 


their 
That 


OF  THE  PARABOLA. 


629 


That  is,  Gi  r  ih  : 
For,  through  the  points 
9,  I,  D,  H,  draw  the  diame- 
ters GK,  iL,  DM,  HN  ;  as 
also  the  lines  ci,  ei,  which 
are  doahle  ordinates  to  the 
diameters  gk,  hn,  by  cor.  1 
theor.  16  ;  therefore 
the  diameters  gk,  bm,  hn, 
bisect  the  lines  cl,  ce,  le  ; 
kence  km  =  cm  —  ck  =  ^ce  —  ^cl  =,4le  =  ln  or  ne, 

and  MN  =  ME    —  NE  -—  iCE  —  iLE  =  ^CL  =  CK  Or  KL. 

But,  by  parallels,  gi  :  ih  :  :  kl  :  ln, 
and         -         -      cg  :  gd  :  :  CK  :  km, 
also         -         -      DH  :  HE  :  :  MN  :  ne. 
But  the  3d  terms  ki,  ck,  mn  are  all  equal ; 

as  also  the  4th  terms    ln,  km,  ne. 
Therefore   the  first  and  second  terms,  in  all  the  lines,  are 
proportional,  namely  gi  :  ih  :  :  cg  :  gd  :  :  dh  :  he.       q.  e.  d. 


THEOREM  XXII. 


If  a  Rectangle  be  described  about  a  Parabola,  having  the 
same  Base  and  Altitude  ;  and  a  diagonal  Line  be  drawn 
from  the  Vertex  to  the  Extremity  of  the  Base  of  the  Para- 
bola,'forming  a  right-angled  Triangle,  of  the  same  Base 
and  Altitude  also  ;  then  any  Line  or  Ordinate  drawn  across 
the  three  Figures,  perpendicular  to  the  Axis,  will  be  cut  m 
Continual  Proportion  by  the  Sides  of  those  Figures. 

A, I       I> 

That  is, 

ep  :  eg  :  :  eg   :  eh. 
Or,  EF,  eg,  eh,  are  in  con- 
tinued proportion. 


For,  by  theor.  1 ,         ab  :  ae  : 
and,  by  sim.  tri.  -     ab   :  ae  : 

theref.  of  equality,     -     ep  :  bc  : 
that  is        -         -         -     EF  :  EH  : 
theref.  by  Geom.  th.  78,  ef,  eg,  eh  are  proportionals, 
•r         •    -         -         -       EP  :  EG  :  ;  eg  :  eh.  q,. 


V©B.  r. 


68 


THBORJEJMt 


53t  CONIC  SECTIONS, 


THEOREM  XXm. 

The  Area  or  Space  of  a  Parabola,  is  equal  to  Two-Thirds  of 
its  Circumscribing  Parallelogram. 

That  is,  the  space  abcga  =  |  abcd  ; 
or  the  space  adcga  =  i  abcd. 
^OR,  conceive  the  space  adcga  to  be  composed  of,  or 
divided  into,  indetiuitely  small  parts,  by  lines  parallel  to 
DC  or  AB,  such  as  ig,  which  divide  ad  into  like  small  and 
equal  parts,  the  number  or  sum  of  which  is  expressed  by  the 
line  AD.     Then, 


by  the  parabc 

•la,         Bcs 

:  eg2    : 

:   AB   ; 

:  AE, 

that  is, 

AD2 

:  Ai2     : 

:  DC   ; 

:  IG. 

Hence  it  follows,  that  any  one  of  these  narrow  parts,  as 
ic 
IG,  is  = X  Ai2  ;    hencej    ad   and   dc    being   given  or 

AD3 

constant  quantities,  it  appears  that  the  said  parts  ig,  &c  are 
proportional  to  ai^,  &c.  or  proportional  to  a  series  of  square 
numbers,  whofee  roots  are  in  arithmetical  progression,  and  the 

DC 

area  adcga  equal  to drawn  into  the  sum  of  such  a  seriqs 

AD2 

of  arithmeticals,  the  number  of  which  is  expressed  by  ad. 

Now,  by  the  remark  at  pag.  227  this  vol.  the  sum  of  the 
squares  of  such  a  series  of  arithmeticals,  is  expressed  by 
^n  ..  w-fl.  2n-j-l»  where  n  denotes  the  number  of  them. 
In  the  present  case,  n  represents  an  infinite  number,  and 
then  the  two  factors  n  -f  1,  2n  -|-  1,  become  only  n  and  2n, 
omitting  the  1  as  inconsiderable  in  respect  of  the  infinite 
number  n :  hence  the  expression  above  becomes  barely 
|n  .  »  .  2w  =  in^. 

To  apply  this  to  the  case  d^ove  :  n  will  denote  ad  or  bc  ; 
and  the  sum  of  all  the  ai^'s  becomes  i  ad^  or  i  bc^  ;  conse- 

DC  DC 

quently  the  sum  of  all  the X  Ai2's,is X    |    ad^  = 

AD2  Ad3 

l  AD  DC  =  i  bd,  which  is  the  area  of  the  exterior  part  adcga.^ 
That  is,  the  said  exterior  part  adcga,  is  i  of  the  parallelo- 
gram ABCD  :  and  consequently  the  interior  part  abcga  is  f 
if  the  same  parallelogram.  9.-  e.  d. 

Corel* 


OF  THE  PARABOLA.  531 

Coral.  The  part  afcga,  inclosed  between  the  cnrve  and 
the  right  liae  afc,  is  ^  of  the  same  parall^lograoi,  bcina;  the 
difference  between  abcga  ami  the  triangle  abcfa,  that  is  be« 
tween  |  aad  i  of  the  parallelugram. 

THEOREM  XXIV. 

The  Solid  Content  of  a  Paraboloid  (or  Solid  generated  by 
the  Rotation  of  a  Parabola  about  its  Axis),  is  equal  to  Half 
its  Circumscribing  Cylinder. 

Let  ABC  be  a  paraboloid,  generated  by  the  rotation  of  the 
parabola  ac  about  its  axis  ad.  ,  Suppose  the  axis  ad  be  divided 
into  an  infinite  number  of  equal  parts,  throu2;h  which  let  cir- 
cular planes  pass,  as  efg,  all  those  circles  making  up  the  whole 
solid  paraboloid. 

Now  if  c  =  the  number 
3-1416,  then  2c  X  fg  is  the 
circumference  of  the  circle 
BFG  whose  radius  is  fg  ;  there- 
fore c  X  fg2  is  the  area  of 
that  circle. 

But,  by  cor.  theor.  1,  Parabola p  X  af  =  fg*,  where  p 
denotes  the  parameter  of  tke  parabola  ;  consequently  pc  X  a? 
will  also  express  the  same  circular  section  eg,  and  therefore 
pc  X  the  sum  of  all  the  af's  will  be  the  sum  of  all  those 
circular  sections,  or  the  whole  content  of  the  solid,  para- 
boloid. 

But  all  the  af's  foroi  an  arithmetical  progression,  begin- 
ning at  0  or  nothing,  and  having  the  greatest  t<^rm  and  the 
sun^  of  all  the  terras  each  expressed  by  the  whole  axis  ad. 
And  since  the  sum  of  all  the  terras  of  such  a  prog^ression,  is 
equal  to  5  AD  X  ad  or  i  ad^  half  the  product  of  the  greatest 
term  and  the  number  of  terms  ;  therefore  i  ad^  is  equal  to 
the  sum  of  all  the  af's,  and  consequently  pc  X  i  ad^,  or  ^  c 
X  /?  X  ad2,  is  the  sum  of  all  the  circular  sections,  or  the 
eontent  of  the  paraboliod. 

DC? 

But,  by  the  parabola,  p  :  dc  : :  dc  :  ad,  orp  = ;  con.- 

AD 

sequently  ^  c  X  p  ^^  ad^  becomes  i  c  X  ^d  X  dc^  for  the 
solid  content  of  the  paraboloid.  B  it  c  X  ad  X  dc^  is  equal 
to  the  cylinder  bcih  ;  consequently  the  paraboloid  is  the  half 
of  its  circumscribing  cylinder.  ^.  k.  s. 


THBORKBJE 


5S2 


CONIC  SECTIONS. 


THEOREM  XXV. 

The  Solidity  of  the  Frustum  begc  of  the  Paraboloid,  is  equal 
to   a   Cylinder  whose  Height  is  df,  and  its  Base  Half  the 
Sum  of  the  two  Circular  Bases  eg,  bc. 
For,  by  the  last  theor.  ^pc  X  ad^  =  the  solid  abc, 
and,  by  the  same,  z  P^  ^  ^^^  ~  ^^®  ^^^^^  '*^^^» 

theref  the  diflf,     ,  i/'c  X  (ad2  —AF2)=the  frust.  begc. 

But  AD2  AF2    =  DF    X(aD   -{-  Af), 

theref  -k  P^  ^  o*   ^  (ad  4-  AF)=the  frust  begc. 

But,  by  the  parab.  p  X  ad  =  rJc^  ,  and  p  X  af  =  fgs  j 
theref.  i  c  X  df  X  (dc2  -j-  fg2)  =  the  frust  begc. 

Q.    E.    ©, 

ON  THE  CONIC  SECTIONS    AS  EXPRESSED  BY   ALGEBRAIC 
EQUATIONS,  CALLED  THE  EQUATIONS  OF  THE  CURVE. 


1    For  the  Ellipse, 

Let  d  denote  ab,  the  transverse,  or  any  diameter  ; 

c  =  iH  its  conjugate  ; 

X  =  AK,  any  absciss,  from  the  extremity  of  the  diam. 

y  =3  DK  the  correspondent  ordirtate. 
Then,  theor.  2,  ab^  :  hi^    :  :  ak  .  kb  :  dk^, 
that  is,  d^  :  c^    :  :  x  [d^ x)  :  2/^, hence  d^  y^  =  c^  {dar—x^)^ 
or  dy  =^  c  ^  {dx  —  x^),  the  equation  of  the  curve. 

And  from  these  equations,  any  one  of  the  four  letters  or 
quantities,  </,  c,  x,  y,  may  easily  be  found,  by  the  reduction  of 
equations,  when  the  other  three  are  given. 

Or,  if/?  denote  the  parameter,  =  c-  -^  </  by  its  definition  ; 
then,  by  cor.  th.  2,  d  :p  :  :  x  (</— jr)  :  y^,or  dy^=p(dx^x^), 
which  is  another  form  of  the  equation  of  the  curve. 


Otherwise* 


EQUATIONS  OF  THE  CURVE.  583 

Otherwise, 


.    Or,  if  d  =  AC  the  semiaxis  ;  c  =  ch  the  semiconjugate  j, 
p  =•  c3  -j-  d  the  gemiparameter  ;  x  =  ck  the  absciss  counted 
irom  the  centre  ;  and  y  =  vrn.  the  ordinate  as  before 
Then  is  ak  =  d-'X,  and  kb=c/  +  ^y  and  ak  .  kb  =  (d^x)  X 

(rf  +   X)   =  d2    _    p2.. 

Then,byth.  2,d2  :c^  ::d^  ^x^iy^  ,2indd^y^  =c^{d2  ^x^), 
OT  dy  =  c  ^  {d^  — ^^),  the  equation  of  the  curve. 

Or,  d  :  p  :  :  d^  —  x^  :  y^,  and  dy^  =  p{d^  —  x^),  another 
form  of  the  equation  to  the  curve  ;  from  which  any  one  of  the 
«[uantities  may  be  found,  when  the  rest  are  given. 

2.  For  the  Hyperbola. 

Because  the  general  property  of  the  opposite  hyperbolas, 
with  respect  to  their  abscisses  and  ordinates,  is  the  same  as 
that  of  the  eUipse,  therefore  the  process  here  is  the  very  same 
as  in  the  former  case  for  the  ellipse  ;  and  the  equation  to  the 
curve  must  come  out  the  same  also,  with  sometimes  only  the 
change  of  the  sign  of  a  letter  or  term  from  -{-  to  — ,  or  from 
—  to  -|->  because  here  the  abscisses  lie  beyond  or  without  the 
transverse  diameter,  whereas  they  lie  between  or  upon  them 
in  the  ellipse.  Thus,  making  the  same  notation  for  the  whole 
diameter,  conjugate,  absciss,  and  ordinate;  as  at  first  in  the  el- 
lipse ;  then,  the  one  absciss  ak  being  x,  the  other  bk  will  be 
d  -h  JT,  which  in  the  ellipse  was  d—x;  so  the  sign  of  x  most 
be  changed  in  the  general  property  and  equation,  by  which  it 
becomes  d^  :  c^  :  :  x  (^d-}-  x)  :  y^  :  hence  d^y-  =  c^  (dx  -\-  x^) 
and  dy  =  c  ^  (^dx  -{-  x^),  the  equation  of  the  curve. 

Or,  usingp  the  parameter  as  before,  it  is,  d  :  p  :  :  x  (d-\-  x): 
y^f  or  dy^  ^=  p  {dx  '\-  x^),  another  form  of  the  equation  to 
the  curve. 

Otherwise^  by  using  the  same  letters  c?,  c,  jo,  for  the  halves 
of  the  diameters  and  parameter,  and  x  for  the  absciss  ck 
counted  from  the  centre  ;  then  is  ak  =  x — rf,  and  bk  =  x-\-d, 
and  the  property  d^  :  c^  :  : .  (x — d)  y.  {x  -{■  d)  :  y^,  j^ives 
d2y2  =  c3  (a;2  —  d^),  or  dy^c^{x^  — d'^),  where  the  signs 
of  d^  and  x^  are  changed  from  what  they  were  in  the  ellipse. 

Or  again,  using  the  semiparameter,  d  :  p  :  :  x^  —  d^  :  y^^ 
and  dy^  =  p  (x^  —  rfa^  the  equation  of  the  curve. 

But  for  the  conjugate  hyperbola,  as  in  the  figure  to  theo- 
rem 3,  the  signs  of  both  x^  and  d^  will  be  positive  ;  for  the 
property  in  that  theorem  being  ca^  ;  cos  :  -.  cd^  -f  ca^  .-  T>e^^ 
'  ''  it 


«34  come  SECTIONS. 

it  is  ds  :  cs  : .-  x^  -\- d^  :  y^  =  j>e^ ,  or  d^y^  =c3  {x^  +  d^),  and 
dy  =^  c  y/  {x^  -\-  rf^),  the  equation  to  the  conjugate  hyperbola. 
Or,  as  c^  :  p  :  :  a;3  -\.  d^  :  y^ ,  and  dy"  =  p  {x^  -f  d^),  also 
the  equation  to  the  same  curve. 


On  the  Equation  to  the  Hyperbola  between  the  Asymptotes. 

\ 


Let  CE  and  cb  be  the  two  as}?mptotes  to 
the  hyperbola  dro,  its  vertex  being  f,  and 
EF,  bd,  AF,  BD,  ordinates  parallel  to  the 
asymptotes.  Put  af  or  ef  =  a.  cb  =  x, 
and  BD  =  y.  Then,  by  theor  28,  af  ef 
=  ee  .  BD,  or  a^  =  xy,  the  equation  to  the 
hyperbola,  when  the  abscisses  and  ordinates 
are  taken  parallel  to  the  asymptotes. 

3.  For  the  Parabola* 


C    b  AB 


If  x  denote  any  absciss  beginning  at  the  verteir,  and  y  its 
ordinate,  also  p  the  parameter.  Then,  by  cor.  theorem  1, 
AK  :  KD  :  :  KD  :  p,  or  x  :  y  :  :  y  :  p  i  hence  px  =  y^  is  th^ 
i^uation  to  the  parabola. 

4.  For  the  Circle. 

Because  the  circle  is  only  a  species  of  the  ellipse,  in  which 
the  two  axes  are  equal  to  each  other  ;  therefore,  making  the 
two  diameters  d  and  c  equal  in  the  foregoing  equations  to  the 
ellipse,  they  become  i/^  =.  dx  —  x^ ,  when  the  absciss  a:  beg^ins 
at  the  vertex  of  the  diameter  :  and  y^  =  d^-^x^,  when  the 
absciss  begins  at  the  centre. 

Scholium. 

In  every  one  of  these  equations,  we  perceive  that  they  rise 
to  the  2d  or  quadratic  degree,  or  to  two  dimensions  ;  which  is 
also  the  number  of  points  in  which  every  one  of  these  curves 
may  be  cut  by  a  right  line.  Hence  it  is  also  that  these  four 
curves  are  said  to  be  lines  of  the  2d  order.  And  these  four 
are  all  the  lines  that  are  of  that  order,  ev^^ry  other  curve  be- 
ing of  some  higher,  or  having  some  higLer  equation,  or  may 
he  cut  in  more  points  by  a  right  line. 


TELEMENTS 


t  63S1 


fiLEMENTS  OF  ISOPERIMETRY. 


I>cf.  1.  When  a  rariable  quantity  has  its  mutations  regu- 
lated  by  a  certain  law,  or  confined  within  certain  limits,  it  ii 
called  a  maximum  when  it  has  reached  the  greatest  magni- 
tude it  can  possibly  attain  ;  and,  on  the  contrary,  when  it  has 
arrived  at  the  least  possible  magnitude,  it  is  called  a  minimum, 

Def.  2.  Isoperimeters,  or  Isoperimetrical  figures j  are  th«se 
which  have  equal  perimeters. 

Def.  3.  The  Locus  of  any  point,  or  intersection,  &c.  is 
the  right  line  or  curve  in  which  these  are  always  situated. 

The  problem  in  which  it  is  required  to  find,  among  figures 
of  the  same  or  of  different  kinds,  those  which  within  equal 
perimeters,  shall  comprehend  the  greatest  surfaces,  has  long 
engaged  the  attention  of  mathematicians.  Since  the  admira- 
ble invention  of  the  method  of  Fluxions,  this  problem  has 
been  elegantly  treated  by  some  of  the  writers  on  that  branch 
of  analysis  ;  especially  by  Maclaurin  and  Simpson.  A  much 
more  extensive  problem  was  investigated  at  the  time  of  "  the 
war  of  problems,"  between  the  two  brothers  John  and  James 
Bernoulli  :  namely,  "  To  find,  among  all  the  isoperimetri- 
cal curves  between  given  limits,  such  a  curve,  that  construct- 
ing a  second  curve,  the  ordinates  of  which  shall  be  functions 
of  the  ordinates  or  arcs  of  the  former,  the  area  oi  the  se- 
cond curve  shall  be  a  maximum  or  a  minimum."  While,  how- 
ever, the  attention  of  mathematicians  was  drawn  to  the  most 
abstruse  inquiries  connected  with  isoperimetry  the  elements  of 
the  subject  were  lost  sight  of  Simpson  was  the  first  who  call- 
ed them  back  to  this  interesting  branch  of  research,  by  giving 
in  his  neat  little  book  of  Geometry  a  chapter  on  the  maxima 
and  minima  of  geometrical  quantities,  and  some  of  the  sim- 
plest problems  concerning  isoperimeters.  The  next  whcr 
treated  this  subject  in  an  elementary  manner  was  Simon  Lhuil- 
lier,  of  Geneva,  who  in  1782,  published  his  treatise  De  Rela- 
tione  mutua  Capacitatis  et  Terminorum  Figu^arum^  &c.  His 
principal  object  in  the  composition  of  that  work  was  to  supply 
the  deficiency  in  this  respect  which  he  found  in  most  of  the 
elementary  Courses,  and  to  determine,  with  regard  to  both, 
the  most  usual  surfaces  and  solids,  those  which  possessed  the 
minimum  of  contour  with  the  same  capacity  ;  and,  recipro- 
cally, the  maximum  of  capacity  with  the  same  boundary.  M. 
Legendre  has  also  considered  the  same  subject  in  a  manner 
somewhat  different  from  either  Simpson  or  Lhuiilier,  in  his 
JBUsnents  dt  Qeomitrie.     Aa  elegant  geonaetriGal  tract,  on  the. 


536  ELEMENTS  OF  ISOPERIMETRT. 

same  subject,  was  also  given,  by  Dr.  Horsley,  in  the  Philo«. 
Trans,  vol.  76,  for  1775  ;  contained  also  in  the  New  Abridg- 
ment, vol  13,  page  653.  The  chief  propositions  deduced  by 
these  four  geometers,  together  with  a  few  additional  proposi- 
tions, are  reduced  into  one  system  in  the  following  theorems. 


SECTION  I.    SURFACES. 

THEOREM  I. 

Of  all  Triangles  of  the  same  Base,  and  whose  Vertices  fall 
in  a  right  Line  given  in  Position,  the  one  whose  Perimeter 
is  a  Minimum  is  that  whose  sides  are  equally  inclined  t* 
that  Line. 

Let  AB  be  the  common  base  of  a  series  of  triangles  abc*. 
ABC,  &c.  whose  vertices  c',  c,  fall  in  the  right  line  lm,  given 
in  position,  then  is  the  triangle  of  least 
perimeter  that  whose  sides  ac,  bc,  are  q, 

inclined  to  the  line  lm  in  equal  angles.  w    ' 


For,  let  BM  be  drawn  from  b,  per- 
pendicularly to  LxM,  and  produced  till 
DM  =  BM  :  join  AD,  and  from  the  point 
c  where  ad  cuts  lm  draw  bc  :  also,  from  any  other  point  cV 
assumed  in  lm,  draw  c  a,  c'b,  c'd.  Then  the  triangles  dmc, 
BMC,  having  the  angle  dcm  =  angle  acl  (th.  7  Geom.)  = 
mob  (by  hyp.)  dmc  =  bmc,  and  dm  =  bm,  and  mc  commoa 
to  both,  have  also  dc  =  bc  ("th.  1  Geom.). 

So  also,  we  have  c'b  =  c  b.  Hence  ac  -f-  cb  =  ac  -|-  cb 
paB  AD,  is  less  than  ac'  +  c'd  (theor.  10  Geom.),  or  than  its 
equal  ac'  +  c'b.  And  consequently,  ab  -f-  bc  +  ac  is  lesi 
than  AB  4*  Bc'  +  Ac'.  q    e.  d. 

Cor.  1.  Of  all  triangles  of  the  same  base  and  the  same  al- 
titude, or  of  all  equal  triangles  of  the  same  base,  the  isoscele§ 
triangle  has  the  smallest  perimeter. 

For,  the  locus  of  the  vertices  of  all  triangles  of  the  same 
altitude  will  be  a  right  line  lm  parallel  to  the  base  ;  and  when 
lm  in  the  above  figure  becomes  parallel  to  ab,  smce  mcb  == 
ACL,  MCB  =  cba  (th.  12  Geom.),  acl  =  cab  ;  it  follow* 
that  CAB  =  cba,  and  consequently  ac  =  cb    (th.  4  Geom.). 

Cor.  2.  Of  all  triangles  of  the  same  surface,  that  which 
Kas  the  minimum  perimeter  is  equilateraL 

For 


SURFACES.  SSV 

Pop  the  triangle  of  the  smallest  perimeter,  with  the  same 
surface,  must  be  isosceles,  whichever  of  the  sides  be  consider- 
ed as  base  :  therefore,  the  triangle  of  smallest  perimeter  hag 
each  two  or  each  pair  of  its  sid^s  equal,  and  consequently  it  is 
^uilateral. 

Cor.  3  Of  all  rectilinear  figures,  with  a  given  magnitude 
and  a  given  number  of  sides,  that  which  has  the  smallest  pe« 
rimeter  is  equilateral. 

For  so  long  as  any  two  adjacent  sides  are  not  equal,  we  may 
draw  a  diagonal  to  become  a  base  to  those  two  sides,  and  then 
draw  an  isosceles  triangle  equal  to  the  triangle  so  cut  off,  but 
«f  less  perimeter  :  whence  the  corollary  is  manifest. 

Scholium. 

To  illustrate  the  second  corollary  above,  the  student  may 
proceed  thus  :  assuming  an  isosceles  triangle  whose  base  is 
not  equal  to  either  of  the  two  sides,  and  then,  taking  for  a  nev9 
base  one  of  those  sides  of  that  triangle,  he  may  construct  an- 
other isosceles  triangle  equal  to  it,  but  a  smaller  perimeter. 
Afterwards,  if  the  base  and  sides  of  this  second  isosceles  tri- 
angle are  not  respectively  equal,  he  may  construct  a  third 
isosceles  triangle  equal  to  it,  but  of  a  still  smaller  perimeter  : 
and  so  ob,  by  performing  these  successive  operations,  he  will 
find  that  all  the  triangles  will  approach  nearer  and  nearec* 
to  an  equilateral  triangle. 


THEOREM  n. 

f)f  all  Triangles  of  the  Same  Base,  and  of  Equal  Perimetert^ 
the  Isosceles  Triangle  has  the  Greatest  Surface. 

Let  ABC,  ABD,  be  two  triangles  of  the  same 
l^ase  AB  and  with  equal  perimeters,  of  which 
the  one  abc  is  isosceles,  the  other  is  not  : 
then  the  triangle  abc  has  a  surface  (or  an 
altitude)  greater  than  the  surface  (or  than 
the  altitude)  of  the  triangle  abd.  

Draw   CD   through    d,    parallel   to   ab,  to    A      E        B 
cut  CE  (drawn  perpendicular  to  ab)  in  c  :  then  it  is  to  be  de- 
monstrated that  CE  is  greater  than  c'e. 

The  triangles  ac  b,  adb,  are  equal  both  in  base  and  altitude  ; 
but  the  triangle  ac'b  is  isosceles,  while  adb  is  scalene  :  there- 
fore the  triangle  ac'b  has  a  smaller  perimeter  than  the  triangle 
ADB  (th.  1  cor.  1),  or  than  acb  (by  hyp.)     Consequently  xff 

Vol.  f.  ea  <Ag  j- 


538  ELEMENTS  OF  ISOPERIMETRY. 

<AC  ;  and  in  the  right-angled  triangles  aec',  aec,  having  ae 
eommon,  we  have  c'e  <  ce*.  q.  e.  i>. 

Cor.  Of  all  isoperirpetrical  figures,  of  which  the  number 
of  sides  is  given,  that  which  is  the  greatest  has  all  its  sides 
equal.  And  in  particular,  of  all  isoperimetrical  triangles,  that 
whose  surface  is  a  maximum,  is  equilateral. 

For,  so  long  as  any  two  adjacent  sides  are  not  equal,  the  sur- 
face may  be  augmented  without  increasing  the  perimeter. 

Remark.  Nearly  as  in  this  theorem  may  it  be  proved 
that,  of  all  triangles  of  equal  heights,  and  of  which  the  sum 
of  the  two  sides  is  equal,  that  which  is  isosceles  has  the  great- 
est base.  And,  of  all  triangles  standing  on  the  same  base 
and  having  equal  vertical  angles,  the  isosceles  one  is  the 
greatest. 

THEOREM  m. 

Of  all  Right  Lines  that  can  be  drawn  through  a  Given  Point, 
between  Two  Right  Lines  Given  in  Position,  that  which  is 
Bisected  by  the  Given  Point  forms  with  the  other  two  Lines 
the  Least  Triangle. 

Of  all  right  lines  gd,  ab,  GD^that 
can  be  drawn  through  a  given  point 
p  to  cut  the  right  lines  ca,  cd,  given 
in  position,  that  ab,  which  is  bisect- 
ed by  the  given  point  p,  forms  with 
CA,  CD,  the  least  triangle,  abc. 

For,  let  EE  be  drawn  through  a       G.%/g-"' 
parallel  to  cd,   meeting   dg  (produ-  E 

ced  if  necessary)  in  e  :  then  the  triangles  pbd,  pae,  are  man- 
ifestly equiangular  ;  and,  since  the  corresponding  sides  pb,  pa 
are  equal,  the  triangles  are  equal  also.  Hence  pbd  will  be 
less  or  greater  than  pag,  according  as  cg  is  greater  or  less 
than  CA.  In  the  former  case,  let  pacd,  which  is  common,  be 
added  to  both  ;  then  will  bac  be  less  than  dgc  (ax.  4  Geom.), 
In  the  latter  case,  if  pgcb  be  added,  dcg  will  be  greater  than 
bag;'  and    consequently   in   this  case  also   bag   is  less  than 

DCG«  Q.    E.    D. 

Cor.  If  PM  and  pn  be  drawn  parallel  to  cb  and  ca  re- 
spectively, the   two  triangles  pam,  pen,  will  be  equal,  and 


*  When  two  ipathematical  quantities  are  separated  by  the  charac- 
ter < .  it  denotes  that  the  preceding  quantity  is  less  than  the  succeed- 
ing one  :  when,  on  the  contrary,  the  sepaj-ating  character  is  >,  it  de- 
notes that  the  preceding  quantity  is  greater  than  the  succeeding  one. 

these 


SURFACES.  639 

these  two  taken  together  (since  am  =  pn  =  mc)  will  be  equal 
to  the  parallelogram  pmcn  :  and  consequently  the  parallelo- 
gram PMCN  is  equal  to  half  abc,  but  less  than  half  dgc. 
From  which  it  follows  (consistently  with  both  the  algebraical 
and  geometrical  solution  of  prob  8,  Application  of  Algebra 
to  Geometry),  that  a  parallelogram  is  always  less  than  half  a 
triangle  in  which  it  is  inscribed,  except  when  the  base  of  the 
one  is  half  the  base  of  the  other,  or  the  height  of  the  former 
half  the  height  of  the  latter  ;  in  which  case  the  parallelogram 
is  just  half  the  triangle  :  this  being  the  maximum  parallelo- 
gram inscribed  in  the  triangle. 

Scholium. 


From  the  preceding  corollary  it  might  easily  be  shown; 
that  the  least  triangle  which  can  possibly  be  described  about, 
and  the  greatest  parallelogram  which  can  be  inscribed  in,  any 
curve  concave  to  its  axis,  will  be  when  the  subtangejnt  is  equal 
to  half  the  base  of  the  triangle,  or  to  the  whole  base  of  the 
parallelogram  :  and  that  the  two  figures  will  be  in  the  ratio  of 
2  to  I.     But  this  is  foreign  to  the  present  enquiry. 

THEOREM  IV. 


Of  all  Triangles  in  which  two  Sides  are  Given  in  Magnitude, 
the  Greatest  is  that  in  which  the  two  Given  Sides  are  Per- 
pendicular to  each  other. 

For,  assuming  for  base  one  of  the  given  sides,  the  surface 
is  proportional  to  the  perpendicular  let  fall  upon  that  side, 
from  the  opposite  extremity  of  the  other  given  side  :  there- 
fore, the  surface  is  the  greatest  when*  that  perpendicular  is 
the  greatest;  that  is  to  say,  when  the  other  side  is  not  in- 
clined to  that  perpendicular,  but  coincides  with  it :  hence  the 
surface  is  a  maximum  when  the  two  given  sides  are  perpendi- 
cular to  each  other. 

Otherwise,  Since  the  surface  of  a  triangle,  in  which  two 
sides  are  given,  is  proportional  to  the  sine  of  the  angle  in- 
cluded between  those  two  sides  ;  it  follows,  that  the  triangle 
is  the  greatest  when  that  sine  is  the  greatest ;  but  the  greatest 
sine  is  the  sine  total,  or  the  sine  of  a  quadrant ;  therefore  the 
two  sides  given  make  a  quadrantal  angle,  or  are  perpendicular 
to  each  ether.  q.  e.  d. 

THBOREAf 


64a  ELEMENTS  OF  ISOPERIMETRY. 


THEOREM  V. 

Of  1^11  Rectilinear  Figures  in  which  all  the  Sides  except  oue 
are  known,  the  Greatest  is  that  which  may  be  Inscribed  in 
a  Semicircle  whose  diameter  is  that  Unknown  Side. 

For,  if  you  suppose  the  contrary  to  be  the  case,  then  when-, 
ever  the  figure  made  with  the  sides  given,  and  the  side  un- 
known is  not  inscribable  in  2l  semicircle  of  which  this  latter 
is  the  diameter,  viz.  whenever  any  one  of  the  angles,  formed 
by  lines  drawn  from  the  extremities  of  the  unknown  side  to 
one  of  the  summits  of  the  figure,  is  not  a  right  angle  ;  we 
may  make  a  figure  greater  than  it,  in  which  that  angle  shall 
be  right,  and  which  shall  only  differ  from  it  in  that  respect : 
therefore,  whenever  all  the  angles  formed  by  right  lines 
drawn  from  the  several  vertices  of  the  figure  to  the  extremi- 
ties of  the  unknown  line,  are  not  right  angles,  or  do  not  fall 
in  the  circumfereace  of  a  semicircle,  the  figure  is  not  in  its 
maximum  state.  ft.  e.  dv 

THEOREM  VI. 

Of  all  Figures  made  with  sides  Given  in  Number  and  Mag- 
nitude, that  which  may  be  Inscribed  in  a  Circle  is  the 
Greatest. 

Let  ABCDEFG  be  the  y 

polygon  inscribed,  and  jS^^zr^  a  ■/^"^"'^^^'^(^ 

abcdefg  a  polygon  with     ^yf^^^^^^^^^^^  / 

equal  sides,  but  not  in-      (  ■     r\        %\  ti- ^^n 

ecribable  in  a  circle  ;  j^tl.......""-- -v  -^i?  \  /^ 

^othatAB=a6,Bc  — 6c,      u^  ij  V 

&c.;  it  is  affirmed  thafe     ^''^i::^-.^,^'^  ^^^„„^b 

the   polygon    abcdepg  <-^  ^ 

is  greater  than  the  polygon  ahcdefg. 

Draw  the  diameter  ep  ;  join  ap,  pb  ;  upon  al?  ^^  ab  make 
the  triangle  abp,  equal  in  all  respects  to  abp  ;  and  join  ep. 
Then,  of  the  two  figures  edcbp^  p(^gf^^  one  at  least  is  not  (by 
hyp  )  inscribable  in  the  semicircle  of  which  ep  is  the  diame- 
ter. Consequently,  one  at  least  of  these  two  figures  is  smaller 
than  the  corresponding  part  of  the  figure  apbcdefg  (th.  5). 
Therefore  the  figure  apdcdefg  is  greater  than  the  figure 
apbcdefg  :  and  if  from  these  there  be  taken  away  the  respec- 
tive triangles  apb,  apbj  which  are  equal  by  construction,  there 
will  remain  (ax  5  Geom.)  the  polygoa  abcdefg  greater  than 
fho  volygon  abcdefg,  q.  e.  d. 

■*    "^  ^  THEOREM 


SURFACES,  041 


THEOREM  Vll. 


*l?he  Magnitude  of  the  Greatest  Polygon  which  can  be  con- 
taiaed  under  any  number  of  Unequal  Sides,  does  not  at  all 
depend  on  the  Order  in  which  thoae  Lines  are  connected 
with  each  other. 

For,  since  the  polygon  is  a  maximum  under  given  sides,  it 
is  inscribable  in  a  circle  (th.  6)  And  this  inscribed  polygon 
is  constituted  of  as  many  isosceles  triangles  as  it  has  sides, 
those  aides  forming  the  bases  of  the  respective  triangles,  the 
other  sides  of  all  the  triangles  being  radii  of  the  circle,  and 
their  common  summit  the  centre  of  the  circle.  Consequently, 
the  magnitude  of  the  polygon,  that  is,  of  the  assemblage  of 
these  triangles,  does  not  at  all  depend  on  their  disposition,  or 
arrangement  round  the  common  centre.  q..  e.  0. 

THEOREM  VHL 

If  a  Polygon  Inscribed  in  a  Circle  have  all  its  Sides  Equal,  all 
its  Angles  are  likewise  Equal,  or  it  is  a  Regular  Polygon. 

For,  if  lines  be  drawn  from  the  several  angles  of  the  poly- 
gon, to  the  centre  of  the  circumscribing  circle,  they  will 
divide  the  polygon  into  as  many  isosceles  triangles  as  it  has 
sides  ;  and  each  of  these  isosceles  triangles  will  be  equal  to 
either  of  the  others  in  all  respects,  and  of  course  they  will 
have  the  angles  at  their  bases  all  equal  :  consequently,  the 
angles  of  the  polygon,  which  are  each  made  up  of  two  angles 
at  the  bases  of  two  contiguous  isosceles  triangles,  will  be  equal 
to  one  another.  q.  e.  d. 

THEOREM  IX. 

Of  all  Figures  having  the  Same  Number  of  Sides  and  Equ  al 
Perimeters,  the  Greatest  is  Regular. 

For,  the  greatest  6gure  under  the  given  conditions  has  all 
>9ides  equal  (th.  2.  cor.).  But  since  the  sum  of  the  sides  and 
the  number  of  them  are  given,  each  of  them  is  given  :  there- 
fore (th.  6),  the  figure  is  inscribable  in  a  circle  :  and  conse- 
quently (th.  8)  all  its  angles  are  equal  ;  that  is,  it  is  regular. 

Q.    E.    D. 

Cor.  Hence  we  see  that  regular  polygons  possess  the  pro- 
perty of  a  maximum  of  surface,  when  compared  with  any 
«Cher  figures  of  the  same  aame  and  with  equal  perimeters. 

THEOREM 


642  ELEMENTS  OF  ISOPERIMETRY. 


THEOREM  X. 

A  Regular  Polygon  has  a  Smaller  Perimeter  than  an  Irregular 
one  Equal  to  it  in  Surface,  and  having  the  Same  Number  of 
Sides. 

This  is  the  converse  of  the  preceding  theorem,  and  may  be 
demonstrated  thus  :  Let  r  and  i  be  two  figures  equal  in  surface 
and  having  the  same  number  of  sides,  of  which  r  is  regular,  i 
irregular  :  let  also  r'  be  a  regular  figure  similar  to  r,  and  hav- 
ing a  perimeter  equal  to  that  of  i.  Then  (th.  9)  r'  >  i  ;  but 
I  ««  r  ;  therefore  r'  >  r.  But  r'  and  a  are  similar  ;  conse- 
quently, perimeter  of  r  >  perimeter  of  r  ;  while  per.  r'  = 
per.  I  {hy  hyp).    Hence,  per.  i  >  per.  r.  q.  e.  d. 

THEOREM  XI. 

The  Surfaces  of  Polygons,  Circumscribed  about  the  Same  or 
Equal  Circles,  are  respectively  as  their  Perimeters*. 

I> 

Let  the  polygon  abcd  be  circumscribed 

about  the  circle  efgh  ;  and  let  this  polygon 
be  divided  into  triangles,  by  lines  drawn 
from  its  several  angles  to  the  centre  o  of 
the  circle.  Then,  since  each  of  the  tan- 
gents, AB,  Bc,  &c.  is  perpendicular  to  its 
corresponding  radius  oe,  of,  &c.  drawn  to  the  point  of  con- 
tact (th.  46  Geom.)  ;  and  since  the  area  of  a  triangle  is  equal 
to  the  rectangle  of  the  i)erpendicular  and  half  the  base  (Mens, 
of  Surfaces,  pr  2)  ;  it  follows,  that  the  area  of  each  of  the 
triangles  abo,  bco,.&c.  is  equal  to  the  rectangle  of  the  radius 
of  the  circle  and  half  the  corresponding  side  ab,  bc,  &c.  :  and 
consequently,  the  area  of  the  polygon  abcd,  circumscribing  the 
circle,  will  be  equal  to  the  rectangle  of  the  radius  of  the  cir- 
cle and  half  the  perimeter  of  the  polygon  But,  the  sur- 
face of  the  circle  is  equal  to  the  rectangle  of  the  radius  and 
half  the  circumference  (th.  94  Geom.).  Therefore,  the  sur- 
face of  the  circle,  is  to  that  of  the  polygon,  as  half  the  cir- 


♦  This  theorem,  togetlier  with  the  anViajrous  ones  respectirg  bodies 
circumscribing"  cyiindevs  and  spheres,  were  given  by  Emerson  in  his 
Geometry,  and  their  use  in  the  themy  of  Isoperimeters  was  justsug- 
geste4  I  but  the  full  appUcation  of  them  to  that  theory  is  due  to  Simon 
Lhuillicr, 

cumference 


SURFACES.  •  b4$ 

cumference  of  the  former,  to  half  the  perimeter  of  the  latter ; 
or,  as  the  circumference  of  the  former,  to  the  perimeter  of 
the  latter.  Now,  let  p  and  p'  be  any  (wo  polygons  circum- 
scribing a  circle  c  :  then,  by  the  foregoing,  we  have 

surf  c  :  surf  p  :  :  circum.  c  :  perim.  p. 

surf  c  :  surf  p'  :  :  circum.  c  :  perim.  p'. 
But,  since  the  antecedents  of  the  ratios  in  both  these  propor- 
tions, are  equal,  the  consequents  are  proportional :  that  is, 
surf,  p  :  surf  p'  :  :  perim.  p  :  perim.  p.  q.  e.  d. 

CoroL  1.  And  one  of  the  triangular  portions  abo,  of  a  po- 
lygon circumscribing  a  circle,  is  to  the  corresponding  circular 
sector,  as  the  side  ab  of  the  polygon,  to  the  arc  of  the  circle 
included  between  ao  and  bo. 

Cor.  2.  Every  circular  arc  is  greater  than  its  chord,  and 
less  than  the  sum  of  the  two  tangents  drawn  from  its  extremi- 
ties and  produced  till  they  meet. 

The  first  part  of  this  corollary  is  evident,  because  a  right 
line  is  the  shortest  distance  between  two  given  points.  The 
second  part  follows  at  once  from  this  proposition  :  for  ea  -\- 
ah  being  to  the  arch  eih,  as  the  quadrangle  aeoh  to  the  cir- 
cular sector  HiEO  ;  and  the  quadr?».ngle  being  greater  than  the 
sector,  because  it  contains  it ;  it  follows  that  ea  -\-  ah  is  great- 
er than  the  arch  eih*. 

Cor,  3.  Hence  also,  any  single  tangeiit  EA,'i3  greater  than 
its  corresponding  arc  EI. 


THEOREM  XII. 

li  a  Circle  and  a  Polygon,  Circumscribable  about  another 
Circle,  are  Isoperimeters,  the  Surface  of  the  Circle  is  a 
Geometrical  Mean  Proportional  between  that  Polygon  and 
a  Similar  Polygon  (regular  or  irregular)  Circumscribed 
about  that  Circle. 

Let  c  be  a  circle,  p  a  polygon  isoperimetrical  to  that  circle, 
and  circumscribable  about  some  other  circle,  and  p'  a  polygon 
similar  to  p  and  circumscribable  about  the  circle  c  ;  it  is  af- 
firmed that  p  :  c  :  :  c  :  p'. 


*  This  second  corollary  is  introduced,  not  because  of  its  immed-ate 
connection  with  the  subject  under  discussion,  but  because,  notwith- 
standingj  its  simplicity,  some  authors  have  employed  whole  pages  in 
attempting  its  demonstration,  and  failed  at  last. 

For, 


544  ELEMENTS  OP  ISOPERIMETRY: 

For,  p  :  !>' :  :  perim^  .  p  :  :  perim^  .  p'  :  :  circum^  .  c  :  p€rW.  p^ 
by  th.  89,  Geom.  and  the  hypothesis 

But  (th.  1 1)  p'  :  c  :  :  per.  p'  :  cir.  c  : :  per^    p'  :  per.  p'Xcir.  c. 

Therefore  p  :  c  :  :  -  -  -  -  cira  .  c  :  per.  pXcir  c. 
:  :  cir.  c  :  per.  p'  :  :  c  :  p'.  .    e.  d. 


THEOREM  Xin. 

U  a  Circle  anci  a  Polygon,  Circumscribable  about  another  Cir- 
cle, are  Equal  in  Surface,  the  Perimeter  of  that  figure  is  a 
Geometrical  Mean  Proportional  between  the  Circumference 
of  the  first  Circle  and  the  Perimeter  of  a  Similar  Polygon 
Circumscribed  about  it. 

Let  c  =  p,  and  let  p'  be  circumscribed  about  c  and  similar 
to  c  :  then  it  is  affirmed  that  cir  c  :  per.  p  :  :  per  p  :  per.  p*- 
For,  cir.  c  :  per.  p  :  :  c  :  p"  :  :  p  :  p'  ;  :  per^  p  :  per^  .  p'. 
Also,  per.  p' :  per  p  -  -  -  -  :  :  per^  p':perpXper  p'« 
Therefore,  cir  c  :  per.  p  -  -  -  :  :  per^.  p  :  per.  X per  t\ 
:  :  per.  p  :  per.  p'.  ^.  b.  Dv. 


THEOREM  XIV. 

The  Circle  is  Greater  than  any  Rectilinear  Figure  of  the  Same 
Perimeter  ;  and  it*  has  a  Perimeter  Smaller  than  any  Recti- 
linear Figure  of  the  Same  Surface. 

For,  in  the  proportion,  p  :  c  :  :  c  :  p',  (th.  12),  since  c  <  p', 

therefore  p  <  c. 
And,  in  the  propor   cir.  c  :  per.  p  :  :  per  p  r  per.  p'  (th.  13)» 
or,  cir.  c  :  per.  p'  :  :  cir^  .  c  :  per^  .  p, 
cir.  c  <  per.  p'  ; 
therefore,  cir  .  c  <  per^  .  p,  or  cir  c  <  per.  p.  q.  e.  ». 

Cor.  1.  It  follows  at  once,  from  this  ancl  the  two  preced- 
ing theorems  that  rectilinear  figures  which  are  isoperimeters» 
and  each  circumscribable  about  a  circle,  are  respectively  in 
the  inverse  ratio  of  the  perimeters,  or  of  the  surfaces,  of 
figures  similar  to  them,  and  both  circumscribed  about  one 
and  the  same  circle.  And  that  the  perimeters  of  equal  rec- 
tilineal figures,  each  circumscribable  about  a  circle,  are  re- 
spectively in  the  subduplicate  ratio  of  the  perimeters  or  of 
the  surfaces,  of  figures,  similar  to  them,  and  both  circumscribed 
about  one  and  the  same  circle. 

Cor.  2.  Therefore,  the  comparisson  of  the  perimeters  of 
ei][ual  regular  figures,  having  different  numbers  g£  sides,  and 

ttiat 


SURFACES.  546 

that  of  the  surfaces  of  regular  isoperimetrical  figures,  is  re- 
duced to  the  comparison  of  the  perimeters,  or  of  the  surfaces 
of  regular  figures  respectively  similar  to  them,  and  circum- 
scribable  about  one  and  the  same  circle. 


Lemma  1. 

If  an  acute  angle  of  a  right-angled  triangle  be  divided  into 
any  number  of  equal  parts,  the  side  of  the  triangle  opposite 
to  that  acute  angle  is  divided  into  unequal  parts,  which  are 
greater  as  they  are  more  remote  from  the  right  angle. 

Let  the  acute  angle  c,  of  the  right- 
angled  triangle  acf,  be  divided  into  equal 
parts,  by  the  lines  gb,  cd,  ce,  drawn  from 
that  angle  to  the  opposite  side  ;  then  shall 
the  parts  ab,  bd,  &c.  intercepted  by  the  A  tjt) 
lines  drawn  from  c,  be  successively  longer 
as  they  are  more  remote  from  the  right  angle  a. 

For  the  angles  acd,  bce,  &c.  beisg  bisected  by  cb,  cd, 
&c.  therefore  by  theor.  83  Geom.  ac  :  cd  :  :  ab  :  bd,  and 
Bc  :  CE  :  :  BD  :  de,  and  dc  :  cf  :  :  de  :  ef.  And  by  th.  21 
6reom.  CD>  ca,  ce>  cb,  cf  >  cd,  and  so  on  :  whence  it 
follows,  that  db>  ab,  de>  db,  and  so  on.  Q.  e-  d. 

Cor.  Hence  it  is  obvious  that,  if  the  part  the  most  remote 
from  the  right  angle  a,  be  repeated  a  number  of  times  equal 
to  that  into  which  the  acute  angle  is  divided,  there  will  re- 
sult a  quantity  greater  than  the  side  opposite  to  the  divided 
angle. 

THEOREM  XV. 

If  two  Regular  Figures,  Circumscribed  about  the  Sam6  Circle, 
differ  in  their  Number  ot  Sides  by  Unity,  that  which  has 
the  Greatest  number  of  Sides  shall  have  the  Smallest  Pe- 
rimeter. 

Let  CA  be  the  radius  of  a  circle,  and  ab,  ad,  the  half  sides 
•f  two  regular  polygons  circumscribed  about  that  circle,  of 
which  the  number  of  sides  differ  by  unity,  being  C^ 
respectively  n  -j-  1  and  n.    The  angles  acb,  acd, 

therefore  are  respectively  the  ^7  and  the  n  th 

part  of  two  right  angles  :    consequentl}?  these       /^     JB  1> 

angles  are  as  n  and   n   -|-    1  :  and  hence,  the  angle  may   be 

conceived  divided  into  n  '\-  \  equal  parts,  ef  which  bod  is  one. 

Vo7.  I.  70  «on- 


546  ELEMENTS  OF  ISOPERIMETRY. 

Consequently,  (cor.  to  the  lemma)  (n  +  I)  bd>  ad.     Taking, 
then,  unequal  quantities  from  equal  quantities  we  shall  have 

(n4"l)  AD  — (w+   1)  BD<(n  -f-  1)  AD  — AD, 
or,  (w  -f-   I)  AB<n  .  AD. 
That  is,  the  semiperimeter  of  the  polygon  whose  half  side  is 
AB  is  smaller  than  the  semiperimeter  of  the  polygon  whose 
half  side  is  ad  :  whence  tho  proposition  is  manifest. 

Cor.  Hence,  augmenting  successively  by  unity  the  num- 
ber of  sides,  it  follows  generally,  that  the  perimeters  of 
polygons  circumscribed  about  any  proposed  circle,  become 
smaller  as  the  number  of  their  sides  become  greater. 

THEOREM  XVI. 

The  Surfaces  of  Regular  Isoperimetrical  Figures  are  Greater 
as  the  Number  of  their  Sides  is  Greater  :  and  the  Perime- 
ters of  Equal  Regular  Figures  are  Smaller  as  the  Number 
of  their  Sides  is  Greater. 

For,  1st.  Regular  isoperimetrical  figures  are  (cor.  1.  th.  14) 
in  the  inverse  ratio  of  fignres  similar  to  them  circumscribed 
about  the  same  circle.  And  (th.  15)  these  latter  are  smaller 
when  the  number  of  sides  is  greater  :  therefore,  on  the 
contrary,  the  former  become  greater  as  they  have  more  sides. 
2dly.  The  perimeters  of  equal  regular  figures  are  (cor.  1 
th.  14)  in  the  subduplicate  ratio  of  the  perimeters  of  similar 
•  figures  circumscribed  about  the  same  circle  :  and  (th.  15) 
these  latter  are  smaller  as  they  have  more  sides  :  therefore 
the  perimeters  of  the  former  also  are  smaller  when  the  num- 
ber of  their  sides  is  greater.  q.  e.  d. 


SECTION  11.  SOLIDS. 

THEOREM  XYH. 

Of  all  Prisms  of  the  Same  Altitude,  whose  Base  is  Given  in 
Magnitude  and  Species,  or  Figure,  or  Shape,  the  Right 
Prism  has  the  Smallest  Surface. 

For,  the  area  of  each  face  of  the  prism  is  proportional  to 
its  height  ;  therefore  the  area  of  each  face  is  the  smallest 
when  its  height  is  the  smallest,  that  is  to  say,  when  it  is  equal 
to  the  altitude  of  the  prism  itself  :  and  in  that  case  the  prism 
is  evidently  a  right  prism.  q.  e,  d. 

THEOREM 


SOLIDS.  547 


THEOREM  XVIII. 

Of  all  Prisms  whose  Base  is  Given  in  Magnitude  and  Species, 
and  whose  Lateral  Surface  is  the  same,  the  Right  Prism  has 
the  Greatest  Ahitude,  or  the  Greatest  Capacity 
This  is  the  converse  of  the  preceding  theorem,  and  may 

readily  be  proved  after  the  manner  of  theorem  2: 

THEOREM  XIX. 

Of  all  Right  Prisms  of  the  Same  Altitude,  whose  Bases  are 
Given  in  Magnitude  and  of  a  Given  numher  of  Sides,  that 
whose  Base  is  a  Regular  Figure  has  the  Smallest  Surface. 

For,  the  surface  of  a  right  prism  of  given  altitude,  and  base 
given  in  magnitude,  is  evidently  proportional  to  the  perime- 
ter of  its  base  But  th  10)  the  base  being  given  in  magni- 
tude, and  having  a  given  number  of  sides,  its  perimeter  is 
smallest  when  it  is  regular  :  whence,  the  truth  of  the  propo- 
sition is  manifest. 

THEOREM  XX. 

Of  two  Right  Prisms  of  the  Same  Altitude,  and  with  Irregu- 
lar Bases  Equal  in  Surface,  that  whose  Base  has  the  Grtiatest 
Number  of  sides  has  the  smallest  Surface  :  and,  in  particu- 
lar, the  Right  Cylinder  has  a  Smaller  Surface  than  any  Prism 
,of  the  Same  Altitude  and  the  Same  Capacity. 

The  demonstration  is  analogous  to  that  of  the  preceding 
theorem,  being  at  once  deducible  from  theorems  16  and  14. 

THEOREM  XXI. 

Of  all  Right  Prisms  whosi  Altitudes  and  whose  Whole  Sur- 
faces are  Equal,  and  whose  Bases  have  a  Given  Number  of 
Sides,  that  whose  Base  is  a  Regular  Figure  is  the  Greatest. 

Let  p,  p',  be  two  right  prisms  of  the  same  name,  equal  in 
altitude,  and  equal  whole  surface,  the  first  of  these  having  a 
regular,  the  second  an  irregular  base  ;  then  is  the  base  of  the 
prism  p,  less  than  the  base  of  the  prism  p'. 

For,  let  p''  be  a  prism  of  equal  altitude,  and  whose  base  is 
equal  to  that  of  the  prism  p'  and  similar  to  that  of  the  prism  p. 
Then  the  lateral  surface  of  the  prism  p"  is  smaller  than  the 
lateral  surface  of  the  prism  p'  (th.  19)  :  hence,  the  total  sur- 
face 


648  ELEMENTS  OF  ISOFERIMETRV. 

face  of  p '  is  smaller  than  the  total  surface  of  p',  and  therefore 
(by  hyp.)  smaller  than  the  whole  surface  of  p.  But  the  prisms 
y  and  p  have  equal  altitudes  and  similar  bases  ;  therefore  the 
dimensions  of  the  base  of  p"  are  smaller  than  the  dimensions 
of  the  base  of  p.  Consequently  the  base  of  p",  or  that  of  p', 
is  less  than  the  base  of  p  ;  or  the  base  of  p  greater  than  that 
of  p  .  q,.  E.  D. 

THEOREM  XXII. 

Of  Two  Right  Prisms,  having  Equal  Altitudes,  Equal  Total 
Surfaces,  and  Regular  Bases,  that  whose  Base  has  the 
Greatest  number  of  Sides,  has  the  Greatest  Capacity.  And, 
in  particular,  a  right  Cylinder  is  Greater  than  any  Right 
Prism  of  Equal  Altitude  and  Equal  Total  Surface. 

The  demonstration  of  this  is  similar  to  that  of  the  preceding 
theorem,  and  flows,  from  th.  20. 

THEOREM  XXm. 

The  Greatest  Parallelopiped  which  can  be  contained  under 
the  Three  Parts  of  a  Given  Line,  any  way  taken,  will  be 
that  constituted  of  Equal  length,  breadth,  and  depth. 

For,  let  AB  be  the  given  line,  and, 

if  possible,  let  two  parts  ae,  ed,  be  [    ■[ 1 '■ — 

unequal.  Bisect  ad  in  c,  then  will  A  '  C  E  I)  B 
the  rectangle  under  ae  (=  ac  -j-  ce) 

and  ED  (=  AC  —  ce),  be  less  than  ac^  ,  or  than  ac  .  cd,  by  the 
square  of  ce  (th.  33  Geom.).  Consequently,  the  solid  ae  . 
ED  .  DB,  will  be  less  than  the  solid  ac  .  cd  .  db  ;  which  is  re-; 
pugnant  to  the  hypothesis. 

Cor.  Hence,  of  all  the  rectangular  parallelopipeds,  having 
the  sum  of  their  three  dimensions  the  same,  the  cube  is  the 
greatest.  •::      '  ' 

THEOREM  XXIV. 

The  Greatest  Parallelopiped  that  can  possibly  be  contained 
under  the  Square  of  one  Part  of  a  Given  Line,  and  the 
other  Part,  any  way  taken,  will  be  when  the  former  Part  Is 
the  Double  of  the  latter. 


Let  AB  be  a  given  line,  and 


I     !■ 


AC  =  2cB,  then  is  ac^  .  cb  the        ~     »      •  / 1         ijiT 
gif-eatest  possible.  A       D  D     C  C        B 

For, 


SOLIDS.  549 

For,  let  Ac'  and  c'b  be  any  other  parts  into  which  the  given 
line  AB  may  be  divided  ;  and  let  ac,  ac',  be  bisected  in  d,  d', 
respectively.  Then  shall  ac^  .  cb  =  4ad  .  dc  .  cb  (cor.  to 
theor.  31  Geom.)  >  4ad'  .  d'c  .  cb,  or  greater  than  its  equal 
c'a^  .  c'b,  by  the  preceding  theorem. 

THEOREM  XXV. 

Of  all  Right  Parallelopipeds  Given  in  Magnitude,  that  which 
has  the  Smallest  Surface  has  all  its  Faces  Squares,  or  is  a 
Cube.  And  reciprocally,  of  all  parallelopipeds  of  Equal 
Surface,  the  Greatest  is  a  Cube. 

For,  by  theorems  19  and  21,  the  right  parallelepiped  hav- 
ing the  smallest  surface  with  the  same  capacity,  or  the  great- 
est capacity  with  the  same  surface,  has  a  square  for  its  base. 
But,  any  face  whatever  may  be  taken  for  base  :  therefore,  in 
the  parallelepiped  whose  surface  is  the  smallest  with  the  same 
capacity,  or  whose  capacity  is  the  greatest  with  the  same  sur- 
face, any  two  opposite  faces  whatever  are  squares  :  conse- 
quently, this  parallelepiped  is  a  cube. 

THEOREM  XXVI. 

The  Capacities  of  Prisms  Circumscribing  the  Same  Right  Cy- 
linder, are  Respectively  as  their  Surfaces,  whether  Total 
or  Lateral. 

For,  the  capacities  are  respectively  as  the  bases  of  the 
prisms  ;  that  is  to  say  (th.  11),  as  the  perimeters  of  their 
bases  ;  and  these  are  manifestly  as  the  lateral  surfaces  :  whence 
the  proposition  is  evident. 

Cor.  The  surface  of  a  right  prism  circumscribing  a  cylin- 
der, is  to  the  surface  of  that  cylinder,  as  the  capacity  of  the 
former,  to  the  capacity  of  the  latter. 

Def.  The  Archimedean  cylinder  is  that  which  circum- 
scribes a  sphere,  or  whose  altitude  is  equal  to  the  diameter  of 
its  base. 

THEOREM  XXVn. 

The  Archimedean  Cylinder  has  a  Smaller  Surface  than  any 
other  Right  Cylinder  of  Equal  Capacity  ;  and  it  is  Greater 
than  any  other  Right  Cylinder  of  Equal  Surface. 

Let  c  andc'  denote  two  right  cylinder?,  of  which  the  first  is 
Archimedean,  the  other  not :  then, 

Ut, 


«65     .        ELEMENfTS  OF  ISOPERIMETRY. 

1st,  If  .  .     c  ==  c',  surf,  c  <  surf,  c  : 
2dly,  if  surf,  c  =  surf,  c',  c  >  c. 

For  having  circumscribed  about  the  cylinders,  c,  c^  tht 
right  prisms  p,  p',  with  square  bases,  the  former  will  be  a 
cube,  the  second  not  :  and  the  following  series  of  equal  ra- 
tios will  obtain,  viz,  c  :  p  :  :  surf,  c  :  surf,  p  ;  :  base  c  :  base  p  :  ; 
base  c'  :  base  p'  :  :  c'  :  p'  :  :  surf,  c'  :  surf.  p'. 

Then,  1st  :  when  c  =  c.  Smce  c  :  p  :  :  c'  :  p',  it  follows 
that  p  =  p  ;  and  therefore  (th.  25)  surf,  p  <  surf.  p.  But, 
surf,  c  :  surf,  p  :  :  surf,  c'  :  surf,  p  ;  consequently  surf.  c< 
surf.  c'.  Q  E.  Id. 

2dly  :  when  surf,  c  =  surf.  c'.  Then,  since  surf,  c  :  surf, 
f  :  :  surf,  c  :  surf,  p',  it  follows  that  surf,  p  =  surf,  p'  ;  and 
therefore  (th.  25)  p  >  p'.  But  c  :  p  :  :  c'  :  p'  ;  consequently 
G>c.  0..  E.  2i>, 

THEOREM  XXVIII. 

Of  all  Right  Prisms  whose  Bases  are  Circumscribable  a\)out 
Circles,  and  Giv€n  in  Species,  that  whose  Altitude  is 
Double  the  Radius  of  the  Circle  Inscribed  in  the  Base, 
has  the  Smallest  Surface  with  the  Same  Capacity,  and  the 
Greatest  Capacity  with  the  Same  Surface. 

This  may  be  demonstrated  exactly  as  the  preceding  theo- 
rem, by  supposing  cylinders  inscribed  in  the  prisms. 

Scholium, 

If  the  base  cannot  be  circumscribed  about  a  circle,  the  right 
prism  which  has  the  minimum  surface  or  the  maximum  capa- 
city, is  that  whose  lateral  surface  is  quadruple  of  the  surface 
of  one  end,  or  that  whose  lateral  surface  is  two-thirds  of  the 
total  surface  This  is  manifestly  the  case  with  the  Archime- 
dean cylinder  ;  and  the  extension  of  the  property  depends 
solely  on  the  mutual  connexion  subsisting  between  the  proper* 
ties  of  the  cylinder,  and  those  of  circumscribing  prisms. 

THEOREM  XXIX. 

The  Surfaces  of  Right  Cones  Circumscribed  about  a  Sphere, 
are  as  their  Solidities. 

For,  it  may  be  demonstrated,  in  a  manner  ianalogous  to 
the  demonstrations  of  theorems  11  and  26,  that  these  cones 

are 


SOLIDS.  55t 

are  equal  to  right  cones  whose  altitude  is  equal  to  the  radius 
of  the  inscribed  sphere,  and  whose  bases  are  equal  to  the 
total  surfaces  of  the  cones  :  therefore  the  surfaces  and  solidi- 
ties are  proportional. 

THEOREM  XXX. 

The  Surface  or  the  Solidity  of  a  Right  Cone  Circumscribed 
about  a  Sphere,  is  Directly  as  the  Square  of  the  Cone'i 
Altitude,  and  Inversely  as  the  Excess  of  that  Altitude  over 
the  Diameter  of  the  Sphere. 
Let  VAT  be  a  right-angled  triangle  which, 
by  its  rotation  upon  va  as  an  axis,  generates 
a  right  cone  ;  and  bda  the  semicircle  which 
by  a  like  rotation  upon  va  forms  the  inscrib- 
ed sphere:  then,  the  surface  or  the  solidity  of 

VA2        * 

the  cone  varies  as . 

VB 

For,  draw  the  radius  cd  to  the  point  of  contact  of  the 
semicircle  at  vt.  Then,  because  the  triangles  vat,  vdc,  are 
similar,  it  is  at  :  vt  :  :  cd  :  vc. 

And,  by  compos,  at  :  at  +  vt  :  cd  :  cd  -|-  cv  =  va  ; 
Therefore  at^  :  (at  +  vt)  at  :  :  cd  :  va,  by  multiply- 

ing the  terms  of  the  first  ratio  by  at. 
But,  because  vb.  vd,  va  are  continued  proportionals, 
it  is  VB  :  VA  :  :  vd^  :  va^  :  :  cd^  :  at^  by  sim.  triangles. 
But  CD  :  VA  :  :  at^  :  (at  -f*  vt)  at  "by  l»he  last  ;  and  these 
mult,  give  cd  .  vb  :  va^  :  :  cd^  :  (at  -|-  ¥t)  at, 

or  VB  :  CD  :  :  va^  :  (at  -|-  vt)  at  =  cd  . . 

^  ^  VB 

But  the  surface  of  the  cone,  which  is  denoted  by  «•  .  at^-J- 
«•.  AT  .  vt*,  is  manifestly  proportional  to  the  first  member 
©f  this  equation,  is  also  proportional  to  the  second  member, 

cr,  since  cd  is  constant,  it  is  proportional  to ,  or  to  a  third 

proportional  to  bv  and  av.  And,  since  the  capacities  of  these 
circumscribing  cones  are  as  their  surfaces  (th.  29),  the  truth 
#f  the  whole  proposition  is  evident. 
Lemma  2. 
The  difiference  of  two  right  lines  being  given,  the  third 
proportional  to  the  less  and  the  greater  of  them  is  a  minimum 
when  the  greater  of  those  lines  is  double  the  other. 


*  jr  being  ^  3'Ul593' 

Let 


552  ELEMENTS  OP  ISOPERIMETRY. 


3         V 


Let  Av  and  bv  be  two  right 
lines,  whose  diflference  ab  is- 
given,  and  let  ap  be  a  third 
proportional  to  bv  and  av  ; 
then  is  ap  a  minimum  when  av  =  2bv. 

For,  since     ap  :  av  :  :  aV  :  bv  ; 

By  division  ap  :  ap — av   :  :  av  :  av — bv  ; 

That  is,         AP  :  vp  :  :  av  ;  ab. 

Hence  vp  .  av  =  ap  .  ab. 
But  vp  .  AV  is  either  =  or   <^Ap2  (cor.  to  th.  31  Geom> 
and  th.  23  of  this  chapter.) 

Therefore  ap  .  AB<iAp3  :  whence  4ab<ap,  or  ap  >  4ab. 
Consequently,  the  minimum  value  of  ap  is  the  quadruple  of 
ab  ;  and  in  that  case  pv  =  va  =  2ab.  ft.  e.  d.* 

THEOREM  XXXI. 

Of  all  Right  Cones  Circumscribed  about  the  Same  Sphere, 
the  Smallest  is  that  whose  Altitude  is  Double  the  Diame- 
ter of  the  Sphere. 

va2 
For,  by  th.  30,  the  solidity  varies  as — (see  the  fig.  t® 

vb 
that  theorem)  :   an^,  by  lemma  2,  since  va  —  vb  is  given,  the 

va2 
third  proportional  —  is  a  minimum  when  va  =  2ab.       q.  e.  d. 

VB 

Cor.  1.  Hence,  the  distance  from  the  centre  of  the  sphere 
to  the  vertex  of  the  least  circumscribing  cone,  is  triple  the 
radius  of  the  sphere. 

Cor.  2.  Hence  also,  the  side  of  such  cone  is  triple  the  radius 
of  its  base. 

♦  Though  the  evidence  of  a  single  demonstration,  conducted  oa 
sound  mathematical  principles,  is  really  irresistible,  and  therefore  needs 
no  corroboration ;  yet  it  is  frequently  conducive  as  well  to  mental  im- 
provement, as  to  mental  delight,  to  obtain  like  results  from  different 
processes.  In  this  view  it  will  be  advantageous  to  the  student,  to  con- 
firm the  truth  of  several  of  the  propositions  in  this  chapter  by  means  of 
the  fluxional  analysis.  Let  the  truth  enunciated  in  the  ab  ve  lemma  be 
taken  for  an  example:  and  let  AB  be  denoted  by  a,  av  by  a:,  b  v  by  x — a^ 

Then  we  shall  have  x— a  :x  ::x  : :  the  third  proportional ;  which 

X — a 
is  to  be  a  minimum.     Hence,  the  fluxion  of  this  fraction  will  be  equal 

x^x  —  2axx 

to  zero  (Flux.  art.  51).     That  is  (Flux.  arts.  19  and  30), «= 

(x-.a)2 
a.    Consequently,  x^  —  2ax  «  o,  and  x  »2a>  or  av  =2ab     ■^    h.ve. 
^         '  THEOREM 


SOLIDS.  663 

THEOREM  XXXIL 

The  Whole  Surface  of  a  Right  Cone  being  Given,,  the  In- 
scribed Sphere  is  the  Greatest  when  the  Slant  Side  of  the 
Cone  is  Triple  the  Kddius  pf  its  Base. 
For,  let   c  and  c'   be  two  right  cones  of  equal  whole  sur- 
face, the  radii  of  their  respective  inscribed  spheres  being 
denoted  by  r  and  r'  ;    let  the  side  of  the  cone  c   be  triple 
the  radius  of  its  base,  the  same  ratio   not  obtaining  in  c'  ; 
and  let  c''  be  a  cone  similar  to  c,  and  circumscribed  about 
the  same  sphere  with  c'.     Then,  (by  th.  31)  surf,  c"  <surf.  c' ; 
therefore  surf   c"  <surf.  c.     But  c"  and  c  are  similar,  there- 
fore all  the  dimensions  of  c"  are  less  than  the  corresponding 
dimensions  of  c  :  and  consequently  the  radius  r'  of  the  sphere 
inscribed  in  c"  or  in  c',  is  less  than  the  radius  r  of  the  sphere 
inscribed  in  c,  or  r  >  r'.  q.  e.  d. 

Cor.  T\ie  capacity  of  a  right  cone  being  given,  the  in- 
scribed sphere  is  the  greatest  when  the  side  of  the  cone  is 
triple  the  radius  of  its  base. 

For  the  capacities  of  such  cones  vary  as  their  surfaces 
{th.  29). 

THEOREM  XXXUI. 

Of  all  Right  Cones  of  Equal  Whole  Surface,  the  Greatest 
is  that  whose  side  is  Triple  the  Radius  of  its  Base  ;  and 
reciprocally,  of  all  Right  Cones  of  Equal  Capacity,  that 
whose  Side  is  Triple  the  Radius  of  its  Base  has  the  Least 
Surface. 

For,  by  th.  29,  the  capacity  of  a  right  cone  is  in  the  com- 
pound ratio  of  its  whole  surface  and  the  radius  of  its  inscribed 
sphere.  Therefore,  the  whole  surface  being  given,  the  ca- 
pacity is  proportional  to  the  radius  of  the  inscribed  sphere  : 
and  consequently  is  a  maximum  when  the  radius  of  the  in- 
scribed sphere  is  such  :  that  is,  (th.  32)  when  the  side  of  the 
cone  is  triple  the  radius  of  the  base*. 

.     Again, 


♦  Here  again  a  similar  result  may  easily  be  dedi'.ced  from  the  method 
of  fluxions.  Let  the  radius  of  the  base  be  denoted  by  x,  the  slant  side 
of  the  cone  by  Zs  its  whole  surface  by  <j2 ,  and  3'14159j  by  ?r.  Then 
the  circumference  of  the  cone*s  basft  will  bfe  2wa:,  its  area  ttx^  and  the 
convex  surface  ttxz.  The  whole  surface  io,  tlierefore,  =  Trx^-^Trxz : 
aa 

and  this  being  s=  as ,  we  have  ^  = ar.    But  the  altitude  of  the 

•ttx 
Vol.  I.  71  cone 


664  ELEMENTS  OF  ISOPERIMETRY. 

Again,  reciprocally,  the  capacity  being  given,  the  surface 
is  in  the  inverse  ratio  of  the  sphere  inscribed  :  \thereforc,  it 
is  the  smallest  when  that  radius  is  the  greatest  ;  that  is  (th.  32) 
when  the  side  of,  the  cone  is  triple  the  radius  of  its  base.  q.  e.  d. 

THEOREM  XlXIV. 

The  Surfaces,  whether  Total  or  Lateral,  of  Pyramids  Cir- 
cumscribed about  the  Same  Right  Cone,  are  respectively 
as  their  Solidities.     And,  in   particular,  the  Surface  of  a 
Pyramid  Circumscribed  about  a  Cone,  is  to  the  Surface  of 
that  Cone,  as  the  SoHdity  of  the  Pyramid  is  to  the  Solidity 
of  the  Cone  ;  and  these  Ratios  are  Equal  to  those  of  the 
Surfaces  or  the  Perimeters  of  the  Bases. 
For,  the  capacities  of  the  several  solids  are  respectively  as 
their  bases  ;  and  their  surfaces  are  as  the  perimeters  of  those 
bases  :    so  that  the   proposition   may   manifestly  be  demon- 
strated by  a  chain  of  reasoning  exactly  like  that  adopted  in 
theorem  11. 


cone  is  equal  to  the  square  root  of  the  difference  of  the  squares  of  the 

a*        2a2 

side  and  of  the  radius  of  the  base  ;  that  is,  it  is  =  v^  (■ ), 

7r2x^      sr 

And  this  multiplied  into  ^  of  the  area  of  the  base,  viz.  by  i'»'*3,give» 

fl4        2a» 
iiTjra  ^  (—  —  — )>  for  the  capacity  of  the  cone.    Now,  this  being 

a  maximum  its  square  must  be  so  likewise  (Flux.  art.  53),  that  is, 

"  •   'f        "      ■",  or,  rejecting  the  denominator,  as  constant,  04x2  — 

9 
27ra2x*  must  be  a  maximum.    This,  in  fluxions,  is  2a  x'x  —  85ra2a:3i 

a2 
=  o  ;  whence  we  have  a*  — 4^Jtr»  =  o,  and  consequently  x=  ^  —  i 

At 
and  fls  =  AiTTX^.    Substituting  this  value  of  a^  for  it,  in  the  value  of  z 
a2  47ra?2 

above  given,  there  results  z  =  —  -  ^c  =        -  —  a:  =»  4«— a?  ■=  Sa^v 

TTX  TtX 

Therefore,  the  side  of  the  cone  is  triple  the  radius  of  its  base.  Or, 
the  square  of  the  altitude  is  to  the  square  of  the  radius  of  the  base^ 
as  8  to  1,  or,  to  the  sqiiare  of  the  diameter  of  the  base,  as  2  to  1. 


THEOREM 


SOLIDS.  555 


THEOREM  XXXV. 

'f^he   Base  of  a  Right  Pyramid   being  Given  in  Species,  the 

Capacity  of  that  Pyramid  is  a  Maximum  with  the  Same 

Surface,  and  on  the  contrary,  the  Surface  is  a  Minimum 

with  the  Same  Capacity,  when  the  Height  of  One  Face  is 

Triple  the  Radius  of  the  Circle  Inscribed  in  the  Base. 

Let  p  and  p'  be  two  right  pyramids  with  similar  bases,  the 

height  of  one  lateral  face  of  p  being  triple  the  radius  of  the 

circle  inscribed  in  the  base,  but  this  proportion  not  obtaining 

jvith  regard  to  p  :  then 

1st.  If  surf.  T  =  surf,  p',  p  >  p' . 
2dly,  if  .  .  p  =  .  .  p',  surf,  p  <  surf  p'. 
For,  let  c  and  c  be  right  cones  inscribed  within  the  pyra- 
mids p  and  p' :  then  in  the  cone  c,  the  slant  side  is  triple 
the  radius  of  its  base,  while  this  is  not  the  case  with  respect 
to  the  cone  c'.  Therefore,  if  c  =  c ,  surf,  c  <  surf,  c'  and 
if  surf  c  =  surf  c',  c  >  c'  (th.  33). 

But,  1st.  surf  p  :  surf  c  :  :  surf,  1p'  :  surf  c' ; 
whence,  if  surf,  p  =  surf,  p  surf,  c  =  surf  c' ; 
therefore  c  >  c'.     But  p  :  c  :  :  p'  :  c'.     Therefore  p  >  p'. 

2dly,  p  :  c  : :  p'  :  c'.  Theref.  if  p=p',  c  =c  :  consequently 
surf  c  <  surf  c'.  But,  surf,  p  :  surf  c  :  :  surf,  p'  :  surf,  c'. 
Whence,  surf,  p  <  surf,  p' 

Q)r.  The  regular  tetraedron  possesses  the  property  of  the 
minimum  surface  with  the  same  capacity,  and  of  the  maxi-. 
mum  capacity  with  the  same  surface,  relatiFely  to  all  right 
pyramids  with  equilateral  triangular  bases,  and,  a  fortiorij 
relatively  to  every  other  triangular  pyramid. 

THEOREM  XXXVI. 

A  Sphere  is  to  any  Circumscribing  Solid,  Bounded  by  Plane 
Surfaces,  as  the  Surface  of  the  Sphere  to  that  of  the  Cir- 
cumscribing Solid. 

For,  since  all  the  planes  touch  the  sphere,  the  radius  drawn 
to  each  point  of  contact  will  be  perpendicular  ta  each  re- 
spective plane.  So  that,  if  planes  be  drawn  through  the  cen- 
tre of  the  sphere  and  through  ail  the  edges  of  the  body,  the 
body  will  be  divided  into  pyramids  whose  bases  are  the  re- 
spective planes,  and  their  common  altitude  the  radius  of  the 
sphere,.  Hence,  the  sum  of  all  these  pyramids,  or  the  whole 
circumscribing  solid,  is  equal  to  a  pyramid  or  a  cone  whose 

base 


656  ELEMENTS  OF  ISOPERIMETRY. 

base  is  equal  to  the  whole  surface  of  that  solid,  and  altitude 
equal  to  the  radius  of  the  sphere.  But  the  capacity  of  the 
sphere  is  equal  to  that  of  a  cone  whose  base  is  equal  to  the 
surface  of  the  sphere,  and  altitude  equal  to  its  radius.  Con- 
sequently, the  capacity  of  the  sphere,  is  to  that  of  the  circum- 
scribing solid,  as  the  surface  of  the  former  to  the  surface  of 
the  latter  :  both  having  in  this  mode  of  considering  them,  a 
common  altitude.  Q,.  e.  d. 

Cor.  1.  All  circumscribing  cylinders,  cones,  &c.  are  to" 
the  sphere  they  circumscribe,  as  their  respective  surfaces. 

For  the  same  proportion  will  subsist  between  their  indefi- 
nitely small  corresponding  segments,  and  therefore  between 
their  wholes. 

Cor,  2.  All  bodies  circumscribing  the  same  sphere,  are 
respectively  as  their  surfaces. 


THEOREM  XXXVU. 


The  Sphere  is  Greater  than  any  Polycdron  of  Equal  Surface. 

For,  first  it  maybe  demonstrated,  by  a  process  similar  to 
that  adopted  in  theorem  9,  that  a  regular  polyedron  has  a 
greater  capacity  than  any  other  polyedron  of  equal  surface. 
Let  p,  therefore,  be  a  regular  polyedron  of  equal  surface  to  . 
a  sphere  s.  Then  p  must  either  circumscribe  s,  or  fall  partly 
within  it  and  partly  out  of  it,  or  fall  entirely  within  it.  The 
first  of  these  suppositions  is  contrary  to  the  hypothesis  of  the 
proposition,  because  in  that  case  the  surface  of  p  could  not 
be  equal  to  that  of  s.  Either  the  2d  or  3d  supposition  there- 
fore must  obtain  ;  and  then  each  plane  of  the  surface  of  p 
must  fall  either  partly  or  wholly  within  the  sphere  s  :  which- 
ever of  these  be  the  case,  the  perpendiculars  demitted  from 
the  centre  of  s  upon  the  planes,  will  be  each  less  than  the 
radius  of  that  sphere  :  and  consequently  the  polyedron  p 
must  be  less  than  the  sphere  s,  because  it  has  an  equal  base, 
but  a  less  altitude.  q.  e.  d. 

Cor.  If  a  prism,  a  cylinder,  a  pyramid,  or  a  cone,  be 
equal  to  a  sphere  either  in  capacity,  or  in  surface  ;  in  the  first 
case,  the  surface  of  the  sphere  is  less  than  the  surface  of  any 
of  those  solids  ;  in  the  second,  the  capacity  of  the  sphere  is 
greater  than  that  of  either  of  those  solids.  ; 

The  theorems  in  this  chapter  will  suggest  a  variety  of 
practical  examples  to  exercise  the  student  in  computation. 
A  few  such  are  given  in  the  following  page. 

EXERCISES. 


SOLIDS.  567 


EXERCISES. 


Ex.  1.  Find  the  areas  of  an  equilateral  triangle,  a  square, 
a  hexagon,  a  dodecagon,  and  a  circle,  the  perimeter  of  each 
being  36. 

Ex.  2.  Find  the  difference  between  the  area  of  a  triangle 
whose  sides  are  3,  4,  and  5,  and  of  an  equilateral  triangle  of 
equal  perimeter. 

Ex.  3.  What  is  the  area  of  the  greatest  triangle  which 
can  be  constituted  with  two  given  sides  8  and  1 1  :  and  what 
will  be  the  length  of  its  third  side  ? 

Ex.  4.  The  circumference  of  a  circle  is  12,  and  the  pe- 
rimeter of  an  irregular  polygon  which  circumscribes  it  is  15: 
what  are  their  respective  areas  ? 

Ex.  5.  Required  the  surface  and  the  solidity  of  the  great- 
est parallelopiped,  whose  length,  breadth,  and  depth,  together 
make  18  ? 

Ex.  6.  The  surface  of  a  square  prism  is  646  :  what  is  its 
solidity  when  a  maximum  ? 

Ex.  7.  The  content  of  a  cylinder  is  169-645968  :  what  is 
its  surface  when  a  minimum  ? 

Ex.  8.  The  whole  surface  of  a  right  cone  is  201-061952  : 
what  is  its  solidity  when  a  maximum  ? 

Ex.  9.  The  surface  of  a  triangular  pyramid  is  43-30127  : 
what  is  its  capacity  when  a  maximum  ?  * 

Ex.  10.  The  radius  of  a  sphere  is  10.  Required  the  so- 
lidities of  this  sphere,  of  its  circumscribed  equilateral  cone, 
and  of  its  circumscribed  cylinder. 

Ex.  11.  The  surface  of  a  sphere  is  28-274337,  and  of  an 
irregular  polyedron  circumscribed  about  it  35  ;  what  are  their 
respective  solidities  ? 

Ex.  12.  The  solidity  of  a  sphere,  equilateral  cone,  and 
Archimedean  cylinder,  are  each  500  :  what  are  the  surfaces 
and  respective  dimensions  of  each  ? 

Ex.  13.  If  the  surface  of  a  sphere  be  represented  by  the 
number  4,  the  circumscribed  cylinder's  convex  surface  and 
whole  surface  will  be  4  and  6,  and  the  circumscribed  equila- 
teral cone's  convex  and  whole  surface,  6  and  9  respectively. 
Show  how  these  numbers  are  deduced. 

Ex.  14.  The  solidity  of  a  sphere,  circumscribed  cylinder, 
and  circumscribed  equilateral  cone,  are  as  the  numbers  4,  6, 
and  9.     Required  the  proof. 


PROBLEMS 


[  658  ] 


J>JiOBLBMS  RELATIVE  TO  THE  DlVISrOlff  OF  FIELDS  ©E 
'  [^.  OTHER  SURFACES. 

PROBLEM  I. 


To  Divide  a  Triangle  into  two  parts  having  a  Given  Ratio, 
m  :  n  " 

1st.  By   a   line  drawn  from  one   angle 
of  the  triangle. 

Make  ad  :  ab  :  :  m  :  m  -{-  n  ;  draw  cd. 
So  shall  ADC,  BDC,  be  the  parts  required. 

m                            n 
Here,  evidently,  ad  = ab,  db  = ab. 

2dly.  By  a  line  parallel  to  one  of  the  sides  of  the  triangle^ 

Let  ABC  be  the  given  triangle,  to  be 
divided  into  two  parts,  in  the  ratio  of  m 
to  n,  by  a  line  parallel  to  the  base  ab. 
Make  ce  to  eb  as  w  to  n:  erect  ed  per- 
pendicularly to  cB,  till  it  meet  the  semi- 
circle described  on  cb,  as  a  diameter,  in 
®D.  Make  cp  =  cd  :  and  draw  through  f,  gf  jj  ab.  So  shall  or 
divide  the  triangle  abc  in  the  given  ratio. 

CD* 

For,  CE  :  cb  = :  :  cd*  (=  cf?)  :  cb^.  ButcEtEP  ::m  :», 

CE 

or  CE  :  cb  :  :  m  :  m  -f-  n,  by  the  construction  :  therefor© 
cf2  :  cb2  :  :in  :tn  -fn.  And  since  A  cgf  :  A  cab  ;  :  CF*  :  cb*  ;, 
it  follows  that  cgf  :  cab  :  :  m  :  m  +  n,  as  required. 

Computation.     Since  cb2  :  cf*    :  :  m  -{-  n  :   w,  therefore, 
(m  +  n)  cf2  =r.  m  .  cb2  ;  whence  cf  ^  (in  -f  »)  =  cb  y'm,. or 

771                                                                      m 
cf  =  cb  ^ ^.-     In  like  manner,  cg  =  ca  ^- . 


mrl;-n 


3dly.  By  a  line  parallel  to  a  given  line. 

Let  HI  be  the  line  parallel  to  which 
a  line  is  to  be  drawn,  so  as  to  divide 
the  triangle  aec  in  the  ratio  of  m 
to  n. 

By  case  2d  draw  gf  parallel  to  ab, 
so  as  to  divide  abc  in  the  given  ratio. 
Through  f  draw  fe  parallel  to  hi. 
On  CE  as  a  diameter  describe  a  semi- 
circle ;  draw  gd  perp.  to  ac,  to  cut 
the  semicircle  in  d.  Make  cp  =  cd  : 
through  p,  parallel  to  ef,  draw  pq,,  the  line  required. 


The 


DIVISION  OF  SURFACES.  659 

The  demonstration  of  this  follows  at  once  from  case  2  ;  be- 
cause it  is  only  to  divide  fce,  by  a  line  parallel  to  fe,  into  two 
triangles  having  the  ratio  of  fce  to  fcg,  that  is,  of  ce  to  cg. 

Computation,  cg  and  cf  being  computed,  as  in  case  I,  the 
distances  ch,  ci  being  given,  and  ct  being  to  cq.  as  ch  to  ci  : 
the  triangles  cgf,  gpq,  also  having  a  common  vertical  angle, 
are  to  each  other,  as  cg  .  cf  to  c^  .  cp.  These  products  there- 
fore are  eqiial  ^  and  since  the  factors  of  the  former  are  known, 
the  latter  product  li  known.  We  have  hence  given  the  ratio 
of  the  two  lines  cp  (=  x)  to  cq,  (—  y)  as  ch  to  ci  ;  say,  asp  to 
q  ;  and   their  product  =  cf  .  cg,  say  =  ab  :  to  find  x  and  y. 

abp  abq 

Here  we  find  x  =  ^  — ,  y  =^  y/  — •     That  is, 
q  P 

CF  .CG  .    CH                                 CF  .  CO  .Cl 
CP  :=  ^  ;    CQ  ==  v/ ' 

CI  CH 

N.  B.  If  the  line  of  division  were  to  be  p»^rpendicular  to 
one  of  the  sides,  as  tocA,  the  construction  would  be  similar  : 

CP  would  be  a  geometrical  mean  between  ca  and —  c&,  6 

being  the  foot  of  the  perpendicular  from  b  upon  ac. 
4thly.     By  a  line  drawn  through  a  given  point  p. 


By  any  of  the  former  cases  draw  Im  (fig*  1)  to  divide  the 
triangle  aibc,  in  the  given  ratio  of  m  to  n  :  bisect  cl  in  r,  and 
through  r  and  m  let  pass  the  sides  of  the  rhomboid  crsm. 
Make  ca  =  pe,  which  is  given,  because  the  point  p  is  given 
in  position  :  make  cd  a  fourth  proportional  to  ca,  cr,  cm  ; 
that  is,  make  ca  :  cr  :  :  cm  :  cd  ;  and  let  a  and  c?,  be  two 
angles  of  the  rhomboid  cahd^  figs.  1  and  2.  pe,  in  figure  2, 
Imping  drawn  parallel  to  ac,  describe  on  ed  as  a  diameter  the 
semicircle  efd,  on  which  set  off  c/*  =  ce  =  ap  :  then  set  off 
d^  or  dn'  on  ca  equal  to  df^  and  through  p  and  m,  p  and  m' 

draw 


560 


DIVISION  OF  SURFACES. 


draw  the  lines  lm,  l'm',  either  of  which  will  divide  the  triangle 
in  the  given  ratio.^ — The  construction  is  given  in  2  figs,  merely 
to  avoid  complexness  in  the  diagrams. 

The  limitations  are  obvious  from  the  construction  :  for,  the 
point  L  must  fall  between  b  and  c,  and  the  point  m  between  a 
and  c  ;  ap  must  also  be  less  than  p6,  otherwise  ef  cannot  be  ap- 
plied to  the  semicircle  on  ed. 

Demon.  Because  cr  =  ic/,  the  rhomboid  crsm  =  triangle! 
elm,  and  because  ca  :  cr  :  :  cm  :  cc/,  we  have  ca  .  cd=cwi  .  cr, 
therefore  rhomboid  cahd  =  rhomboid  crsm  =  triangle  elm. 
By  reason  of  the  parallels  cb,  fed,  and  ca,  ah,  the  triangles 
aLP,  c?GM,  feop,  are  similar,  and  are  to  each  other  as  the 
squares  of  their  homologous  sides  ap,  dM,  fep  :  now  ed^  =  ef^ 
4-cf/'^,  by  construction  ;  and  ed  =  rb,  ef  =  ap,  df  =  dm  ; 
therefore  pfe^  =  ap^  -}-  du^ ,  or,  the  triangle  pfeo  taken  away 
from  the  rhomboid,  is  equal  to  the  sum  of  the  triangles  aPL, 
djAGj  added  to  the  part  capod  :  consequently  clm  =  cafed,  as 
required,  By  a  like  process,  it  may  be  shown  that  aLP,  ^g'm', 
pfeo',  are  similar,  and  «l'p  +  dou  =  pfeo' ;  whence  Fbda'  = 
aL'p,  and  cl'm'  =  caferf,  as  required. 

Computation,  cl,  cm,  being  known,  as  well  as  ca,  ap,  or 
cc,  cp,  cr  =  ^cZ,  is  known  :  and  hence  erf  may  be  found  by 
the  proportion  ca  :  cr  :  :  cm :  cd.  Then  cd  —  cc  =  ed,  and 
^ed^^ej^  =  ^ed2-_ap2  =  d/*  =  ^m  =   ck'.     Thus  cm   is 

C/  .  cm 

determined.     Then  we  have =  cl. 


N.  B.  When  the  point  is  in  one  of  the  sides,  as  at  m 
make  cl  .  cm  .  (m  -f-  n)  =  ca  .  cb  .  m,  or,  cl  :  ca  : :  m 
{m-\-n)  CM,  and  the  thing  is  done. 

5thly.     By  the  shortest  line  possible. 

Draw  any  line  p^  dividing  the  triangle  in 
the  given  ratio,  and  so  that  the  summit  of  the 
triangle  cpq  shall  be  c  the  most  acute  of  the 
three  angles  of  the  triangle.  Make  cm  =  cn, 
a  geometrical  mean  proportional  between  cp 
and  CQ  ;  so  shall  mn  be  the  shortest  line  pos- 
sible dividing  the  triangle  in  the  given  ratio. — 
The  computation  is  evident. 

Demons.  Suppose  mn  to  be  the  shortest 
line  cutting  off  the  given  triangle  cmn,  and 

CG  _£_  MN  .  MN  =  MG  4"  «N  =  CG  .  COt  M  -}- 
CG  .  COt  N  =  CG  (cot  M-J-COt  n).   But,  COt  M-f- 

cos  M     COS  N      sin  (m-^k) 
cot  n  = 


;  then 
.  CB  : 


-+- 


sm  M      sin  N    sm  M  .  sm  n 


And  (equa. 


XVIII, 


DIVISION  OF  SURFACES.  561 

XVIII,  Analj^t.  pi.  trigonom.)  sin  m  .  sin  n  =  Jcos  (m— n)— ^cos 

sin.(M+N) 
(M-|-N)=icos(M— N)-flcosc.Theref.MN=cG. ; 

^CG8(m  —  N)«f-iCv«C 

which  expression  is  a  minimum  when  its  denominator  is  a 
maximum  :  that  is,  When  cos  (m— n)  is  the  greatest  possible, 
which  is  manifestly  when  m — n  =  o,  ofm  =n,  or  when  the 
triangle  cmn  is  isosceles.  That  the  isosceles  triangle  must 
have  the  most  acute  angle  for  its  summit,  is  evident  from  the 
consideration,  that  since  2Acmn  =  cg  .  mn,  mn  varies  in- 
versely as  CG  ;  and  consequently  mn  is  shortest  when  cg  is 
longest,  that  is,  when  the  angle  c  is  the  most  acute. 

N.  B.  A  very  simple  and  elegant  demonstration  to  this 
case  is  given  in  Simpson's  Geometry  :  vide  the  book  on  Max. 
and  Min.  See  also  another  demonstration  at  case  2d  prob, 
6th,  below. 

PROBLEM  II. 

To  Divide  a  Triangle  into  three  Parts,  having  the  Ratio  of 
the  quantities  m,  n,  p. 

1st.  By  lines  drawn  from  one  angle  of  the  triangle  to  the 
opposite  side. 

Divide  the  side  ab,  opposite  the    angle    c  G 

from  whence  the  lines  are  to  proceed,  in  the 
given  ratio  at  d,  e  ;  join  cd,  ce,   and    acd, 

DOE,  ECB,   are  the   three   triangles    required. 

The  demonstration  is  manifest ;  as  is  also  the    ,^.D      E  B 
computation.  .  -^' 

If  it  be  wished  that  the  lines  of  division  be  the  shortest  the 
nature  of  the  case  will  admit  of.  Jet  them  be  drawn  from  the 
most  obtuse  angle,  to  the  opposite  or  longest  side. 

2dly.  By  lines  parallel  to  one  of  the  sides  of  the  triangle.    , 

Make  cd  :  dh  :  hb  :  :  m  :  «  :  p.  Erect 
DE,  HI,  perpendicularly  to  cb,  till  they  meet 
the  semicircle  described  on  the  diameter 
CB  in  E  and  i.  Make  cf  =  ce,  and  ck  = 
ci.     Draw  gf  through  f,  and  lk  through  k, 

parallel  to  ab  ;  so  shall  the  lines  gf  and  lk, 

divide  the  triangle  abc  as  required.  <">  B 

The  demonstration  and  computation  will  be  similar  to  those 
in  the  second  case  of  prob.  1. 

3dly.     By  lines  drawn  from  a  given  poiut  on  one  of  the 
sides. 
Vol.  I.  72  Fig. 


562  DIVISION  OF  SURFACES. 


Fig.  2. 


A  fljypw)  B  A.  a    b  T  B 

Let  p  (fig.  1)  be  the  given  point,  a  and  b  the  points  which 
divide  the  side  ab  in  the  given  ratio  of  m,  n,  p  :  the  point  p 
falling  between  a  and  6.  Join  pc,  parallel  to  which  draw  ac, 
bd,  to  meet  the  sides  ac,  bc  in  the  points  c  and  d  :  join  pc, 
p</,  so  shall  the  lines  cp,  prf,  divide  the  triangle  in  the  given 
ratio. 

In  fig.  2,  where  p  falls  nearer  one  of  the  extremities  of  ab 
than  both  a  and  6,  the  construction  is  essentially  the  same  ; 
the  sole  difference  in  the  result  is,  that  the  points  c,  and  d, 
both  fall  on  one  side  ac  of  the  triangle. 

Demon.  The  lines  ca,  c6,  divide  the  triangle  into  the  given 
ratio,  by  case  1st.  But  by  reason  of  the  parallel  lines  aCy  Pc, 
bd,  A  O.CC  A  ocp,  and  A  bdc  =  bdF,  Therefore,  in  fig  1, 
Aac  -{-  ac?  =  a«c  +  «fc  that  is,  acp  =  Aac  :  and  Bbd  -\-  bdp . 
=  tibd  +  bdc,  that  is,  bc?p  =  b6c.  Consequently,  the  re- 
mainder ccpd  =  cab. — In  fig.  2,  acp  =  Aac,  and  xdv  =  ac6  ; 
therefore  cpd  =  acp  ;  and  acb— A«ifp  =  acb— ac6,  that  is, 
CB?d  =■  chb. 

Computation.  The  perpendiculars  eg,  cd  being  demitted, 
A  ACP  :  A  ACB  : ;  w  :  m-^-n-^rp  :  :  ap  .  c^  :  ab  .  cd.     Therefore 

W . AB . CD 

(m-]rn-\-p)  AP.cg=ni.AB.cD,  and  eg  =  ■ .      The  line 

(»m4"«+/')ap 
eg  being  thus  known,  we  soon   find  ac  ;  for  cd  :  ac  :  :  c^  : 

AC  '  eg  ?w  .  AB  .  AC 

AC  = = .     Indeed  this  expression  may  be 

CD  (mi-H+/')AP 

deduced  more   simply  ;    for,   since   acb  :  acp  :   :  ac  .  ab  ; 
AC  .  AP  :  :  m-\-n-\-p  :  m,  we  have  (m-f-w-f-p)  ac  .  AP=m  .  ab  .  ac, 

m . ab     AC 
and  AC  = .      By  a  like  process  is  obtained,  in 

(m+w4-/))AP 

/J-AB      BC  »n+«)  AB  .  AC 

fig.  1,  Bfi  = ;  and,  in  fig,  2,  Ad  = . 

(ra-|-n+/>)  PB  C^^'+w+A)  AP 

4ihly.  By  lines  drawn  from  a  given  point  p  within  the 
triangle. 


Const, 


J 


DIVISION  OF  SURFACES 
c 


663 


Const.  Through  p  and  c  draw  the  line  cpp,  and  let  the 
triangle  be  divided  into  the  given  ratio  by  lines  pc,  pd^  drawn 
from  p  to  intersect  ac,  bc,  or  either  of  them  ;  according  to 
the  method  described  in  case  3  of  this  problem.  Through  p 
draw  PC,  pc?,  and  respectively  parallel  to  them,  from  p  draw 
the  lines  /?m,  /jn  :  join  pm,  pn  ;  so  shall  these  lines  with  p/), 
divide  the  triangle  in  the  given  ratio. 

Demon.  The  triangles  cpm,  crp^  are  manifestly  equal,  as 
are  also  c?pn,  drp  ;  therefore  cpm  =  cpc,  and  cpn  =  cpd  ; 
whence  also,  in  fig.  1,  cnpm  =:  cdpc,  and,  in  fig.  2,  cb/jpn  = 
CBpd. 

cp    cd 

Compvi.    Since  cp  .  cn  ==  cp  .  cd,  we  have  cn  = • 


In  like  manner  cm  = 


CP 

cp    cc 


Remark.  It  will  generally  be  best  to  contrive  that  the 
smallest  share  of  the  triangle  shall  be  laid  off  nearest  the  ver- 
tex c  of  the  triangle,  in  order  to  ensure  the  possibility  of  the 
construction.  Even  this  precaution  however  may  sometimes 
fail, ♦of  ensuring  the  construction  by  the  method  above  given  : 
when  this  happens,  proceed  thus  : 

By  case  1 ,  draw  the  hues  cc?,  cc,  from 
the  vertex  c  to  the  opposite  side  ab,  to  di- 
vide the  triangle  in  the  given  ratio  Upon 
AB  set  off  any  where  mx,  so  that  mn  :  ab  :  : 
pp  (the  perp.  from  p  on  ab)  :  cp,  the  alti- 
tude of  the  triangle.  U  mp  and  pn  are  to- 
gether to  be  the  least  possible,  then  set  off  i  mn  on  each  side 
the  point  p  :  so  will  the  triangle  mpn  be  isosceles,  and  its 
perimeter  (with  the  given  base  and  area)  a  minimum. 

5thly.  By  lines,  one  of  which  is  drawn yrom  a  given  angle 
to  a  given  point,  which  is  also  the  point  of  concourse  of  the 
other  two  lines. 


Const. 


664  DIVISION  OF  SURFACES. 


e 

A 

4 

^ 

\^a         &  B 

(^omt.  By  case  1st  draw  the  lines  ca,  c6,  dividing  the 
triangle  in  the  given  ratio,  and  so  that  the  smaller  portions 
shall  lie  nearest  the  angles  a  and  b  (unless  the  conditions  of 
the  divison  require  it  to  be  otherwise).  From  p  and  a  demit 
upon  AC  the  perpendiculars  pp,  ac ;  and  from  ?  and  6,  on 
Bc,  the  perpendiculars  pg,  hd.  Make  cm  :  ca  :  :  .ac  :  vp^  and 
CN  :  cB  ;  :  6c?  :  vq.  Draw  pm,  pn,  which,  with  cp,  will  divide 
•the  triangle  as  required. 

When  the  perpendicular  from  h  or  from  a,  upon  bc  or  ac, 
is  longer  than  the  corresponding  perpendicular  from  p,  the 
point  N  or  m  will  fall  further  from  c  than  b  or  a  does.  Sup- 
pose it  to  be  N  :  then  make  n'  e  :  cb  :  :  nc  :  cp,  and  draw  pn' 
for  the  line  of  division. 

The  demonstration  of  all  this  is  too  obvious  to  need  trac- 
ing here. 

CA  .  ac 

Corn-put.     The  perp.  ca  =  Aa  .  sin  a  ;  and  cm  = . 

CB  .  bd 
hd  =  Bb  .  sin  b  ;  and  cn  = . 

Fq 

Gthly.  By  lines,  one  of  which  falls  from  the  given  po^nt  of 
concourse  of  ^11  three,  upon  a  given  side,  in  a  given  angle. 

Suppose  the  given  angle  to  be  a  right 
angle,  and  p/"  the  given  perpendicular  : 
which  will  simplify  the  operation,  though 
the  principles  of  construction  will  be  the 
same.  ^_ 

Const,  Let  ca,  c&,  divide  the  triangle  M  A 
in  the  given  ratio.  Make/N  :  cb  :  :  bd  :  f/, 
and  /m  :  ca  :.:  ac  :  t/'^  and  draw  pn,  pm,  thus  forming  two 
triangles  p/n,  p/m,  equal  to  c6b,  coa,  respectively.  If  n  fall 
between  /  and  b,  and  m  between  a  and  /,  this  construction 
manifestly  effects  the  division.  But  if  one  of  the  points,  sup- 
pose M,  falls  beyond  the  corresponding  point  a,  the  line  pm 
intersecting  ac  in  e  :  then  make  m'e  :  ca  :  :  cm.  :  cp,  and  draw 
pm'  :  so  shall  p/,  pm',  pn,  divide  the  triangle  as  required. 


Compvt, 


DIVISION  OF  SURFACES.  66S 

Comput.     Here  ca  and  bd,  are  found  as  in  case  6th  ;  and 
CB  '  bd  CA  '  ac 

l^ence>  = ;  and/M  = .  ThenPM=:v^(M/3+/p3), 

p/  P/ 

^/ 
and  —  =  sin.  m.  Also  I  H0°  —  (m+a)  »  mca.     Then  sin  m€a  : 

PM 

sin  M  :  sin  a  oc  ma  (=«/'— a/)  :  ac  :  mc.  Again  pe  =  pm-m«  ; 
A'  .  eM 

and  lastly  M'e  = ^- 

ev 
Here  also  the  demonstration  is  manifest. 
7thly.     By  lines  drawn  from  the  angles  to  meet  in  a  deter- 
minate point. 

Construe.  On  one  of  the  sides,  as  ac,  set 
off  AD,  so  that  AD  :  AC  :  :  m  :  m  -\-  n  -{-  p. 
And  on  the  other,  as  ab,  set  off  be,  so  that 

BE  :  Bc  :  :  n  :  m  -{-  n -\-  p.   Through  n  draw . 

DG  parallel  to  ab  ;  and  through  e,  eh  parallel  -A.  IE\  JS 
to  Bc  ;  to  their  poiut  of  intersection  i  draw  lipes  ai,  bi,  ci> 
which  will  divide  the  triangle  abc  into  the  portions  required* 
Demon.  Any  triangle  whose  base  is  ab,  and  whose  vertex 
ialls  in  dg  parallel  to  it,  will  manifestly  be  to  abc,  as  ad  to 
AC,  or  as  m  to  m  +  »*  "f*  p  :  so  also,  any  triangle  whose  base 
is  EC,  and  whose  vertex  falls  in  eh  parallel  to  it,  will  be  to 
ABC,  as  BE  to  ba,  that  is,  as  n  to  m  -f-  ^  4*  p« 

Thus  we  have  aib  :  acb  :  :  m  :  m-|-n+p, 
and     .     .     .     bic  :  acb  :  :  n   :  in-\-n~{-py 
therefore      .     aib  :  bic  :  :  m  :  n. 
And  the  first  two  proportions  give,  by  composition. 
aib  4-  BIC   :  ACB  f:m-|-n   :    m  -\-  n  -^  p  ;  and  by  division, 
acb  — (aib -{-bic)  :  acb  .•*:  w  -j-  n  +  p' —  {m-{-n)  :  m-\-n-\'p, 
or  Aic  :  agb   -.  \  p   :  m  +  n  -^  p,  consequently  aib  :  bic  :  aic 
<JC  m  :  71   :   p. 

n .  AB                      tn  .  BC 
Comput.     be  =  Gi  = ;  bg  = ;    angle   bgi 

=  2  right  angles  —  b.  Hence,  in  the  triangle  bgi,  there  are 
known  two  sides  and  in  the  included  angle,  to  find  the  third 
side  BI. 

C 
Remark.    When  m=-  n  =p,the  constraction  ^//K 

becomes  simpler.     Thus  :  from  the  vertex  draw       -^^^ni^ 
GD  to  bisect  AB  ;  and  from  b  draw  be  in  like      a^-j^    3 
manner  to  the  middle  of  ac  :  the  point  of  inter- 
section I  of  the  lines  cd,  be,  will  be  the  point  sought. 

For,  on  be  and  be  produced,  demit,  from  the  angles  c  and 


566  DIVISION  OF  SURFACES. 

A,  the  perpendiculars  ci,  ak  :  then  the  triangles  cei,  aek, 
are  equal  in  all  respects,  because  ae  =  ce,  kae  =  ice,  and 
the  angles  at  e  are  equal.  Hence  ak  =  ci.  But  these  are 
the  perpendicular  altitudes  of  the  triangles  bp(  ,,  bpa,  which 
have  the  common  base  bp.  Consequently  those  two  triangles 
are  equal  in  area.  In  a  similar  manner  it  may  be  proved,  that 
AFC  =  ape  or  CFB  :  therefore  these  three  triangles  are  equal 
to  each  other,  and  the  lines  pa,  pb,  pc,  trisect  the  A  abc. 

PROBLEM  III. 

To  Divide  a  Triangle  into  Four  Parts,  having  the  Proportion 
of  the  Quantities  m,  n,  p,  q. 

This  like  the  former  problems,  might  be  divided  into  sev- 
eral cases,  the  consideration  of  all  which  would  draw  us  to  a 
very  great  length,  and  which  is  in  a  great  measure  unnecessary, 
because  the  method  will  in  general  hf  suggested  immediately 
on  contemplating  the  method  of  proceeding  in  the  analogous 
case  of  the  preceding  problem.  We  shall  therefore  only  take 
one  case,  namely,  that  in  which  the  lines  of  division  must  all 
be  drawn  from  a  given  point  of  one  of  the  sides. 

Let  p  be  the  given  point  in  the  side  ab. 

Let  the  points,  /,  m,  w,  divide  the  base  ab 
in  the  given  proportion  ;  so  will  the  lines  cZ, 
cm*,  en,  divide  the  surface  of  the  triangle  in 

the  same    proportion.     Join  cp,  and  parallel     ^^ — ^ 

to  it  draw,  from  /,  m,  n,  the  lines  II,  mM,  /iN,  A./  2»  r  ^  J5 
to  cut  the  other  two  sides  of  the  triangle  in  l,  m,  n.  Draw 
PL,  PM,  PN,  which  will  divide  the  triangle  as  required. 

The  demonstration  is  too  obvious  to  need  tracing  through- 
out :  for  the  triangles  l/p,  lZc,  having  the  same  base  l/,  and 
lying  between  the  same  two  parallels  l/,  cp,  are  equal  ;  to 
each  of  these  adding  the  triangle  al/,  there  results  alp  =  ac/. 
And  in  like  manner  the  truth  of  the  whole  construction  may 
be  shown. 

The  computation   may  be  conducted  after  the  manner  of 
that  in  case  3d  prob.  2. 

PROBLEM  IV. 

To  Divide  a  Quadrilateral  into  Two  Parts  having  a  Given 
Ratio,  m  ;  n. 

1st.     By  a  line  drawn  from  any  point  in  the  perimeter  of 
the  figure. 

Construe.  From  p  draw  lines  pa,  pb, 
to  the  opposite  angles  a,  b.  Through  d 
draw  DF  parallel  to  pa,  to  meet  ba  pro- 
duced  in  F  :  and  through  c  draw   ce  pa- 

rallel  to   pb  to  meet  ab  produced  in  e.    c  A       M   B  E 

Divide 


DIVISION  OF  SURFACES.  567* 

Divide  fe  Iq  m,  in  the  given  ratio  of  m  to  n  :  join  p,  m  ;  so 
shall  the  line  pm  divide  the  quadrilateral  as  required. 

Demon.  That  the  triangle  fpe  is  equal  to  the  quadrangle 
ABCD,  may  be  shown  by  the  same  process  as  is  used  to  demon- 
strate the  construction  of  prob.  36,  Geometry  ;  of  which,  in 
fact,  this  is  only  a  modification.  And  the  line  pm  evidently 
divides  fpe  in  the  given  ratio.  But  fpm  :;=  adpm,  and  epm  = 
BCPM  :  therefore  pm  divides  the  quadrangle  also  in  the  given 
ratio. 

Remark  1.  If  the  line  pm  cut  either  of  the  sides  ad,  bc, 
then  its  position  must  be  changed  by  a  process  similar  to  that 
described  in  the  6th  and  6th  cases  of  the  last  problem. 

Remark  2.  The  quadrilateral  may  be  divided  into  three, 
four,  or  more  parts,  by  a  similar  method,  being  subject  how- 
ever to  the  restriction  mentioned  in  the  preceding  remark. 

Remark  3.  The  same  method  may  obviously  be  used  when 
the  given  point  p  is  in  one  ol  the  angles  of  the  figure. 

Comput.  Suppose  i  to  be  the  point  of  intersection  of  the 
sides  DC  and  ab,  produced  ;  and  let  the  part  of  the  quadrila- 
teral laid  off  towards  i,  be  to  the  other,  as  n  to  m.  Then  we 
n(lD  .  lA-lB  .  ic) 

have  iM  = .    As  to  the  distances  di,  ai,  (since 

{m-^n)  IP 
the  angles  at  a  and  d,  and  consequently  that  at  i,  are  known), 
they  areieasily  found  from  the  proportionality  of  the  sides  of 
triangles  to  the  sines  of  their  opposite  angles. 

2dly.  By  a  line  drawn  parallel  to  a  given  line. 

Construe.  Produce  dc,  ab,  till 
they  meet,  as  at  i.  Join  db  pa- 
rallel to  which  draw  cf.  Divide  af 
in  the  given  ratio,  in  h.  Through 
D  draw  DG  parallel  to  the  given 
line.  Make  ip  a  mean  proportional  \jcr^P~H~B" 
between  ih,  ig  ;  through  p  draw 

PM  parallel  to  gd  :  so  shall  pm  divide  the  quadrilateral  abcd 
as  required. 

Demon,  It  is  evident,  from  the  transformation  of  figures, 
so  often  resorted  to  in  these  problems,  that  the  triangle  adf 
=  quadri^teral  abcd  (th.  36  Geom.)  :  and  that  dh  divides 
the  triangle  adf  in  the  given  ratio,  is  evident  from  prob.  1 
case  1.  We  have  only  then  to  demonstrate  that  the  triangle 
iHD  is  equal  to  the  trianstle  ipm,  for  in  that  case  hdf  will 
manifestly  be  equal  to  bcmp.  Now,  by  construction,  ih  : 
IP  :  :  IP  :  IG  :  :  (by  the  parallels)  im  :  id  ;  whence,  by  making 
the  products  of  the   means   and  extremes  equal,   we   have 


568  DIVISION  OF  SURFACES. 

ID  .  iH  =  IP  .  iM  ;  but  when  the  products  of  the  sides  aboui 
the  equal  angles  of  two  triangles  having  a  common  angle  are 
equal,  those  triangles  are  equal  ;  therefore  A  ihd  =  A  ipm. 

Q.  E.  D. 

Comput.  In  the  triangle*  adi,  adg  are  given  all  the  angles, 
and  the  side  ad  ;  whence  ai,  ag,  di,  and  ic,  =  bi — dc,  be- 
come known.  In  the  triangle  ifc,  all  the  angles  and  the  side 
IC  are  known  ;  whence  if  becomes  known,  as  well  as  fh, 
since  ah  :  hf  :  :  m  :  n.  Lastly,  ip  =  -v/(ih  .  ig),  and  ig  : 
ID   :  :  IP  :  im. 

Cor.  1 .  When  the  line  of  division  pm  is  to  be  perpendicu- 
lar to  a  side,  or  parallel  to  a  given  side  ;  we  have  only  to  draw 
DG  accordingly  :  so  that  those  two  cases  are  included  in  this. 

Cor.  2.  When  the  line  pm  is  to  be  the  shortest  possible,  it 
ciust  cut  off  an  isosceles  triangle  towards  the  acutest  angle  ; 
and  in  that  case  ig  must  evidently  be  equal  to  id. 

3dly.  By  a  line  drawn  through  a  given  point. 

The  method  will  be  the  same  as  that  to  case  4th  prob.  1, 
and  therefore  need  not  be  repeated  here. 

Scholium.  If  a  quadrilateral  were  to  be  divided  into  four 
parts  in  a  given  proportion,  m,  n,  p,  g  :  we  must  first  divide  it 
into  two  parts  having  the  ratio  of  ?«  ■+•  w,  to  p  -j-  g  ;  and  thea 
each  of  the  quadrangles  so  forined  into  their  respective  ra- 
tios, of  m  to  w,  and  p  to  gf. 

PROBLEM  V. 

To  Divide  a  Pentagon  into  Two  Parts  having  a  Given  Ratio, 
from  a  Criven  Point  in  one  of  the  Sides. 

Reduce  the  pentagon  to  a  triangle  by  prob.  37,  Geometry, 
and  divide  this  triangle  in  the  given  ratio  by  case  1  prob.  1. 

PROBLEM  VL 

To  Divide  any  Polygon  into  Two  Parts  having  a  Giveii< 
Ratio. 

1st.  From  a  given  point  in  the  perimeter  of  the  polygon. 

Construe.     Join  any   two    opposite  ^~^C 

angles  a,  d,  of  the  polygon  by  the  line  A^^*^^^^ 

ad.     Reduce    the  part  abcd  into  an  ../"     i\\  \\ 

equivalent  triangle  nps,  whose  vertex  •••'/        pV  V'^; 

shall  be  the  given  point  p,  and  base  ad  [         j  Ju  /^^ 

produced  :  an  operation  which  may  be  G^^     j/jj 

performed  at  once,  if  the  portion  abcd  ^  KME 

be  quadrangular  ;  or  by  several  opera- 
tions. 


DIVISION  OF  SURFACES.  569 

tions  (as  from  8  sides  to  6,  from  6  to  4,  &c.)  if  the  sides  be 
more  than  four.  Divide  the  triangle  nps  into  two  parts  bar- 
ing the  given  ratio,  by  the  line  ph.  In  like  manner,  reduce 
ADEFGA  into  an  equivalent  triangle  having  h  for  its  vertex, 
and  FE  produced  for  its  base  ;  and  divide  this  triangle  into  the 
given  ratio  by  a  line  from  h,  as  hk.  The  compound  line  phk 
will  manifestly  divide  the  whole  polygon  into  two  parts  having 
the  given  ratio.  To  reduce  this  to  a  right  line,  join  pk,  and 
through  H  draw  hm  parallel  to  it ;  join  pm  ;  so  will  the  right 
line  PM  divide  the  polygon  as  required,  provided  m  fall  between 
F  and  E.  If  it  do  not,  the  reduction  may  be  completed  by  the 
process  described  in  cases  5th  and  6th  prob.  2d. 

All  this  is  too  evident  to  need  demonstration. 

R&mark  There  is  a  direct  method  of  solving  this  problem, 
without  subdividing  the  figure  :  but  as  it  requires  the  compu- 
tation of  the  area,  it  is  not  given  here. 

2dly.     By  the  shortest  line  possible. 

Construe.  From  any  point  f', 
in  one  of  those  two  sides  of  the 
polygon  which,  when  produced, 
meet  in  the  most  acute  angle  i, 
draw  a  line  p'm',  to  the  other 
of  those  sides  (ef),  dividing  the 
polygon  in  the  giren  ratio.  Find  ^  ^ 

the  points  p  and  m,  so  that  ip  or  im  shall  be  a  mean  propor- 
tional between  ip',  im'  ;  then  will  *m  be  the  line  of  division 
required. 

The  demonstration  of  this  is  the  same  as  has  been  already 
given  at  case  5  prob.    1.     Those,  however,  who  wish  for  a 
proof,  independent  of  the  arithmetic  of  sines,  will  not  be  dis- 
pleased to  have  the  additional  demonstration  below. 
• 

The  shortest  line  which,  with  two  other  lines  given  in  po- 
sition,  includes  a  given  area,  will  make  equal  angles  with  those 
two  lines,  or  with  the  segments  of  them  it  cuts  off  from  an 
isosceles  triangle. 

Let  the  two  triangles  abc,  aef,  having  the  common  angle 
A,  be  equal  in  surface,  and  let  the  former  triangle  be  isosce- 
les, or  have  ab  =  ac  ;  then  is  eg  shorter  than  ef. 

Vol.    I.  73  First, 


670 


DIVISION  OF  SURFACES. 


First,  the  oblique  base  ef  cannot  pass 
through  D,  the  rpiddle  point  of  bc,  as  in 
the  annexed  figure.  For,  drawing  cg 
parallel  to  ab,  to  meet  ef  produced  in 
G.  Then  the  two  triangles  dbe,  dcg 
are  identical,  or  mutually  equal  in  all 
respects.  Consequently  the  triangle 
DCF  is  less  than  dbe,  and  therefore  abc 
less  than  aef.  f 


'  EF  must  therefore  cut  bc  in  some  point  h  between  b  and  d, 
and  cutting  the  perp.  ad  in  some  point  i   above  d,  as  in  the 

2d  fig.  Upon  EF  (produced  if  necessary) 
demit  the  perp.  ak.  Then  in  the  right- 
angled  A  AiK,  the  perp.  ak  is  less  than 
the  hypothenuse  ai,  much  more  then  is  it 
less  than  the  other  perp.  ad.  But^,  of 
equal  triangles,  that  which  has  the  greatest 
perpendicular,  has  the  least  base.  There- 
fore the  base  bc  is  less  than  the  base  ef.  q,  e.  d. 

This  series  of  problems  might  have  been  extended  much 
further  ;  bat  the  preceding  will  furnish  a  sufficient  variety, 
to  suggest  to  the  student  the  best  method  to  be  adopted  in 
almost  any  other  case  that  may  occur.  The  following  practi- 
cal examples  are  subjoined  by  way  of  exercise. 

Ex.  1.  A  triangular  field,  whose  sides  are  20,  18,  and  16 
chains,  is  to  have  a  piece  of  4  acres  in  content  fenced  off  from 
it,  by  a  right  line  drawn  from  the  most  obtuse  angle  to  the 
opposite  side.  Required  the  length  of  the  dividing  line,  and 
its  distance  from  either  extremity  of  the  line  on  which  it 
falls? 

.  Ex.  2.  The  three  sides  of  a  triangle  are  5,  12,  and  13.  If 
two -thirds  of  this  triangle  be  cut  off  by  a  line  drawn  parallel 
to  the  longest  side^  it  is  required  to  find  the  length  of  the 
dividing  line,  and  the  distance  of  its  two  extremities  from  the 
extremities  of  the  longest  side. 

Ex.  3.  It  is  required  fo  find  the  length  and  position  of  the 
shortest  possible  line,  which  shall  divide,  into  two  equal  parts, 
a  tjriangle  whose  sides  are  25,  24,  and  7  respectively. 

Ejt:.'4.  The  sides  of  a  triangle  arc  6,  8,  and  10  :  it  is  re- 
quired to  cut  off  nine-sixteenths  of  it,  by  a  line  that  shall  pass 
through  the  centre  of  its  inscribed  circle. 

Ex. 


DIVISION  OF  SURFACES.  571 

Ex.  S.  Two  sides  of  a  triangle,  which  include  an  angle  of 
70'',  and  14  and  J  7  respectively.  It  is  required  to  divide  it 
into  three  equal  parts,  by  lines  drawn  parallel  to  its  longest 
side. 

Ex,  6.  The  base  of  a  triangle  is  1 1 2*66,  the  vertical  angle 
S7<»  57',  and  the  difference  of  the  sides  about  that  angle  is  8. 
It  is  to  be  divided  into  three  equal  parts,  by  lines  drawn  from 
the  angles  to  meet  in  a  point  within  the  triangle.  The  lengths 
of  those  lines  are  required. 

Ex.  7.  The  legs  of  a  right-angled  triangle  are  28  and  43. 
Required  the  lengths  of  lines  drawn  from  the  middle  of  the 
hypothenuse,  to  divide  it  into  four  equal  parts. 

Ex.  8.  The  length  and  breadth  of  a  rectangle  are  16  and 
9.  it  is  proposed  to  cut  off  one-iifth  of  it,  by  a  line  which 
shall  be  drawn  from  a  point  on  the  longest  side  at  the  dis- 
tance of  4  from  a  corner. 

Ex.  9.  A  regular  hexagon,  each  of  whose  sides  is  12,  is 
to  be  divided  into  four  equal  parts,  by  two  equal  lines  ;  both 
passing  through  the  centre  of  the  figure.  What  is  the  length 
of  those  lines  when  a  minimum  ? 

Ex.  10.  The  three  sides  of  a  triangle  are  5,  6,  and  7.  How 
may  it  be  divided  into  four  equal  parts,  by  two  lines  which 
shall  cut  each  other  perpendicularly  ? 

^*^  The  student  will  find  that  some  of  these  examples  will 
admit  of  two  answers. 


On  the  Construction  of  Geometrical  Problems, 

Problems  in  Plane  Geometry  are  solved  either  by  means  of 
the  modern  or  algebraical  analysis,  or  of  the  ancient  or  geo- 
metrical analysis.  Of  the  former,  some  specimens  are  given 
in  the  Application  of  Algebra  to  Geometry,  page  369,  &;c.  of 
this  volume.  Of  the  latter,  we  here  present  a  few  examples, 
premising  a  brief  account  of  this  kind  of  analysis. 

Geometrical  analysis  is  the  way  by  which  we  proceed  from 
the  thing  demanded,  granted  for  the  moment,  till  we  have 
connected  it  by  a  series  of  consequences  with  something  an- 
teriorly known,  or  placed  it  among  the  number  of  principles 
known  to  be  true,  \ 

Analysis 


572  GONSTRITCTION  OF 

AaslyUB  may  be  distinguished  into  two  kinds.  In  th«  one, 
which  18  named  by  Pappus,  contemplative,  it  is  proposed  to 
ascertain  the  truth  or  the  falsehood  of  a  proposition  advanced; 
the  other  is  referred  to  the  solution  of  problems,  or  to  the 
investigation  of  unknown  truths.  In  the  first  we  assume  as 
true,  or  as  previously  existing,  the  subject  of  the  proposition 
advanced,  and  proceed  by  the  consequences  of  the  hypothesis 
to  something  known  ;  and  if  the  result  thus  found  be  true, 
the  proposition  advanced  is  likewise  true.  The  direct  de- 
monstration is  afterwards  formed,  by  taking  up  again,  in  ^u 
inverted  order,  the  several  parts  of  the  analysis.  If  the  con- 
sequence at  which  we  arrive  in  the  last  place  is  found  fake, 
we  thence  conclude  that  the  proposition  analysed  is  also  false. 
When  a  problem  is  under  consideration,  we  first  suppose  it 
resolved,  and  then  pursue  the  consequences  thence  derived 
till  we  come  to  somethi&g  known.  If  the  ultimate  result 
thus  obtained  be  comprised  in  what  the  geometers  call  data, 
the  question  proposed  may  be  resolved  :  the  demonstration 
(or  rather  the  construction),  is  also  constituted  by  taking  the 
parts  of  the  analysis  in  an  inverted  order.  The  impossibiHty 
of  the  last  result  of  the  analysis,  will  prove  evidently,  in  this 
case  as  well  as  in  the  former,  that  of  the  thing  required. 

In  illustration  of  these  remarks  take  the  following  ex- 
amples. 

Ex.  1.  It  is  required  to  draw,  in  a  given  segment  of  a 
circle,  from  the  extremes  of  the  base  a  and  b,  two  lines  ac, 
Bc,  meeting  at  a  point  c  in  the  circumference,  such  that  they 
shall  have  to  each  other  a  given  ratio,  viz.  that  of  m  to  w. 

Analysis.  Suppose  that  the  thing  is  af- 
fected, that  is  to  say,  that  ac  :  cb  :  :  m  :  n, 
and  let  the  base  ab  of  the  segment  be  cut 
in  the  same  ratio  in  the  point  e.  Then  eg, 
being  drawn,  will  bisect  the  angle  ace  (by 
th.  83Geom.);  consequently,  if  the  circle  'Tf' 

be  completed,  and  ce  be  produced  to  meet  it  in  f,  the  re- 
maining circumference  will  also  be  bisected  in  f,  or  have 
FX  =  FB,  because  those  arcs  are  the  double  measures  of  equal 
angles  :  therefore  the  point  f,  as  well  as  e,  being  given,  the 
point  c  is  also  given. 

Construction.  Let  the  given  base  of  the  segment  Afe  be 
cut  in  the  point  e  in  the  assigned  ratio  of  m  to  n,  and  com- 
plete the  circle  ;  bisect  the  remaining  circumference  in  f  ; 
join  FE,  and  produce  it  till  it  meet  the  circumference  in  c  ; 
then  drawing  ca,  cb,  the  thing  is  done. 

Demonstration.  Since  the  are  fa  =«  the  arc  fb,  the  angle 
ACF  *»  angle  bcf,  by  theor.  49  Geom.  ;  therefore  ac  :  cb  ;  : 

AE  : 


GEOMETRICAL  PROBLEMS. 


573 


Az  :  EB,  by  th.  83.     Bot  ae  :  eb  :  :  m  :  n,  by  constructioa  ; 
therefore  ac  :  cb  :  :  m  :  n.  q.  e.  d. 

Ex.  2.  From  a  give^  circle  to  cut  off  an  arc  such,  that 
the  suQEi  of  m  times  the  sine,  and  n  times  the  versed  sine,  may 
be  equal  to  a  given  lioe. 

Anal.  Suppose  it  done,  and  that  aee'b  is 
the  given  circle,  be'k  the  required  arc,  ed  its 
sine,  BD  its  versed  sine  ;  in  da  (produced  if  ne- 
cessary) take  HP  anwth  part  of  the  given  sum  ; 
join  PE,  and  produce  it  to  meet  bf  4-  to  ab  or  || 
Uf  ED,  in  the  point  p.  Then,  sin^e  m  .  eu  -{■  n 
.  BD  =  '^  .  Bf  =  n  .  Tt)  -\-  n  .  BD  ;  consequently 
m  .  ED  =  n  .  PD  ;  hence  pd  :  ed  ;  :  m  :  Ji  "^ 
(by  sim.  tri.)  pb  :  bf  ;  therefore  pb  :  bf   : 


B 
But  pd 


:  ED    :  : 

Now  PB 

is  given,  therefore  bf  is  given  in  magnitude,  and,  being  at  right 
angles  to  pb,  is  also  given  in  position  ;  therefore  the  point  f  is 
given  and  consequently  pf  given  in  position  ;  and  therefore 
the  point  e,  its  intersection  with  the  circumference  of  the  cir- 
cle aee'b,  or  the  arc  be  is  given.     Hence  the  followiog. 

Const.  From  b,  the  extremity  of  any  diameter  ab  of  the 
given  circle,  draw  em  at  right  angles  to  ab  ;  in  ab  (produced 
i£  necessary)  take  bp  an  nth  part  of  the  given  sum  ;  and  on 
BM  take  bf  so  that  bf  :  bp  :  :  n  :  m.  Join  pf,  meeting  the 
circumference  of  the  circle  in  e  and  e',  and  be  or  be'  is  the  arc 
required. 

Demon.  From  the  points  e  and  e'  draw  ed  and  e'd'  at  right 
angles  to  ab.  Then,  since  bp  :  bp  :  :  ra  :  m,  and  (by  sim.  tri.) 
BP  :  BP  :  :  de  :  dp  ;  therefore  de  :  dp  :  :  /t  :  7/1.  Hence  m  .  de 
=  n  .  DP  ;  add  to  each  n  .  bd,  then  will  m  -  oe  -|-  n  .  bd  =  w  . 
BD  -j-  «  •  DP  ==  w  .  PB,  or  the  given  sum. 

Ex.  3.  In  a  given  triangle  abh,  to  inscribe  another  tri- 
angle abCf  similar  to  a  given  one,  having  one  of  its  sides  pa- 
rallel to  a  line  mBu  given  by  position,  and  the  angular  points 
a,  6,  c,  situate  in  the  sides  ab,  bh,  ah,  of  the  triangle  abh 
respectively. 

Analysis.  Suppose  the  thing  done, 
and  that  ahc  is  inscribed  as  required. 
Through  any  point  c  in  bh  draw  cd 
parallel  to  mzn  or  to  ab,  and  cutting 
AB  in  D  ;  draw  ce  parallel  to  he,  and 
bE  to  ac,  intersecting  each  other  in  e. 
The  triangles  dec,  ac6,  are  similar,  and  dc  :  a6  :  :  ce  :  be  ; 
also  bdc,  Bab,  are  similar,  and  pc  :  a6  :  :  bo  :  b6.     Therefore 


674  CONSTRUCTION  OF,  &c. 

Bc  :  CE  :  :  Bh  ;  be;  and  they  are  about  equal  angles,  conse- 
quently B,  E,  c,  are  in  a  right  line. 

Construe.  From  any  point  c  in  bh,  draw  cd  parallel  to  nm; 
on  CD  constitute  a  triangle  cde  similar  to  the  given  one  ;  and 
through  its  angles  e  draw  be,  which  produce  till  it  cuts  ah  in 
€  :  through  c  draw  ca  parallel  to  ed  and  cb  parallel  to  eg  ;  join 
fl6,  then  abc  is  the  triangle  required,  having  its  side  ab  parallel 
to  mn,  and  being  similar  to  the  given  triangle. 

Demon.  For,  because  of  the  parallel  hnes  ac,  de,  and  cb, 
EC,  the  quadrilaterals  bdec  and  Bac6,  are  similar  ;  and  there- 
fore the  proportional  lines  dc,  a6,  cutting  off  equal  angles 
BDc,  Bab  ;  BCD,  Bba  ;  must  make  the  angles  edc,  ecd,  respec- 
tively equal  to- the  angles  cab.  cha ;  while  ab  is  parallel  to  dc, 
which  is  parallel  to  mBw,  by  construction. 

Ex.  4.  Given,  in  a  plane  triangle,  the  vertical  angle,  the 
perpendicular,  and  the  rectangle  of  the  segments  of  the  base, 
made  by  that  perpendicular  ;  to  construct  the  triangle.  ' 

Anal.  Suppose  abc  the  triangle  re- 
quired, BD  the  given  perpendicular  to  the 
base  AC,  produce  it  to  meet  the  periphery 
©f  the  circumscribing  circle  abch,  whose 
centre  is  o,  in  h  ;  then,  by  th.  61  Geom. 
the  rectangle  bd  .  dh  =  ad  .  dc,  the  given 
rectangle  :  hence,  since  bd  is  given,  dh 
and  BH  are  given  ;  therefore  bi  =  hi  is  given  :  as  also 
id  =  OE  :  and  the  angle  eoc  is  =  abc  the  given  one,  be- 
cause Eoc  is  measured  by  the  arc  kc,  and  abc  by  half  the 
arc  akc  or  by  kc  Consequently  ec  and  ac  =  2ec  are  given. 
Whence  this 

Construction.     Find  dh  such,   that  db  .  dh   =  the   given 

AD  .  DC 

rectangle,  or  find  dh  =  ;    then   on   any   right  line 

BD 

GF  take  FE  =  the  given  perpendicular,  and  eg  ~  dh  ;  bisect 
FG  in  o,  and. make  eoc  =  the  given  vertical  angle  ;  then 
will  oc  cut  EC,  drawn  perpendicular  to  oe,  in  c.  With  cen- 
tre o  and  radius  oc,  describe  a  circle,  cutting  ce  produced  in 
A  :  through  f  parallel  to  ac  draw  fb,  to  cut  the  circle  in  b  ; 
join  AB,  cb,  and  abc  is  the  triangle  required. 

Remark.     In  a  similar  manner  we  may  proceed,  when  it 
is  required  to  divide  a  given  angle  into  two  parts,  the  rect- 
angle 


QUESTIONS  IN  MENSURATION.  57fi 

angle  of  whose  tangents  may  be  of  a  given  magnitude.     Sec 
prob,  40,  Simpson's  Select  Exercises. 

Note.  For  other  exercises,  the  student  may  construct  all 
the  problems,  except  the  24th,  in  the  Application  of  Algebra 
to  Geometry,  at  page  369,  &c.  of  this  volume  And  that  he 
may  be  the  better  able  to  trace  the  relative  advantages  of  the 
ancient  and  the  modern  analysis,  it  will  be  adviseabit  that  he 
solve  those  problems  both  geometrically  and  algebraically. 

• 


PRACTICAL  EXERCISES  IN  MENSURATION. 

Quest.  1.  WHAT  difference  is  there  between  a  floor 
28  feet  long  by  20  broad,  and  two  others,  each  of  half  the 
dimensions  ;  and  what  do  all  three  come  to  at  45s.  per  square, 
or  100  square  feet  ? 

Ans.  diff.  280  sq.  feet     Amount  18  guineas. 

Quest.  2.  An  elm  plank  is  14  feet  3  inches  long,  and  I  would 
have  just  a  square  yard  slit  off  it  ;  at  what  distance  from  the 
edge  must  the  line  be  struck  ?  Ans.  7|i  inohes. 

Quest.  3.  A  ceiling  contains  114  yards  6  feet  of  plastering, 
and  the  room  28  feet  broad  ;  what  is  the  length  of  it  ? 

Ans.  36|  feet. 

Quest.  4.  A  common  joist  is  7  inch©*  deep»  and  2i 
thick  ;  but  wanting,  a  scantling  just  as  big  again  that  shafl 
be  3  inches  thick  ;  what  will  the  other  dimensions  be  ? 

Ans.  11|  inches. 

Quest.  5.  A  wooden  cistern  cost  me  3s.  ^d.  painting 
within,  at  Cd.  per  yard  ;  the  length  of  it  was  102  inches, 
and  the  depth  21  inches  ;  what  was  the  width  ? 

Ans.  27i  inches. 

Quest.  6.  If  my  court-yard  be  47  feet  9  inches  square, 
and  I  have  laid  a  foot-path  with  Purbeck  stone,  of  4  feet 
wide,  along  one  side  of  it,  what  will  paving  the  rest  with 
flints  come  to,  at  6d.  per  square  yard  ?  Ans.  bl.  16s.  OJrf. 

Quest.  7.  A  ladder,  26f  feet  long,  may  be  so  .planted, 
that  it  shall  reach  a  window  22  feet  from  the  ground  on 
one  side  of  the  street*;  and  by  only  turning  it  over,  without 
moving  the  foot  out  of  its  place,   it  will  do  the  same  by  a 

window 


i- 


a76  QUESTIONS  IN  MENSURATION. 

windoiT  14  feet  higk  an  the  other  side  ;  what  is  the  breadth  of 
the  street  ?  Ans.  37  feet  9|  inches. 

QuHST.  8.  The  paving  of  a  triangular  court,  at  18d. 
per  foot,  came  to  100/.  ;  the  longest  of  the  three  sides  waa 
88  feet  ;  required  the  sum  of  the  oth^r  two  equal  sides  ? 

Ans.  106-86  f^et. 

Quest.  9.  There  are  two  columns  in  the  ruins  af  Perse- 
polis  left  standing  upright  :  the  one  is  64  fe|t  above  the 
plain,  and  the  other  60  :  in  a  straight  line  Between  these 
stands  an  ancient  small  statue,  the  head  of  which  is  97  feet 
from  the  summit  of  the  higher,  and  86  feet  from  the  top  of 
the  lower  column,  the  base  of  which  is  just  76  feet  from  the 
statue's  base.  Required  the  distance  between  the  tops  of  the 
two  columns  ?  Ans.  167  feet  nearly. 

Quest.  10.  The  perambulator,  or  surveying  wheel,  is  so 
contrived,  as  to  turn  just  twice  in  the  length  of  1  pole,  or 
16^  feet  ;  required  the  diameter  ?  Ans.  2-626  feet. 

Quest.  11.  In  turning  a  one-horse  chaise  within  a  ring 
of  a  certain  diameter,  it  was  observed  that  the  outer  wheel 
made  two  turns,  while  the  inner  made  but  one  :  the  wheels 
were  both  4  feet  high  ;  and  supposing  them  fixed  at  the  dis- 
tance of  5  feet  asunder  on  the  axletree,  what  was  the  circum- 
ference of  the  track  described  by  the  outer  wheel  ? 

Ans.  62-83  feet. 

Quest.  12.  What  is  the  side  of  that  equilateral  triangle, 
whose  area  cost  as  much  paving  at  8d.  ai  foot,  as  the  palli- 
sading  the  three  sides  did  at  a  guinea  a  yard  ? 

Ans.  72-746  feet. 

Quest.  13.  In  the  trapezium  abcd,  are  given,  ab  =  13, 
Bc  =  31i,  CD  =  24,  and  da  =  18,  also  b  a  right  angle  ;  re- 
quired the  area  ?  Ans.  410-122. 

Quest.  14.  A  roof  which  is  24  feet  8  inches  by  14  feet 
6  inches,  is  to  be  covered  with  lead  at  81b.  per  square  foot  : 
what  will  it  come  to  at  18s.  per  cwt.  ?        Ans.  22/.  19s.  lOirf. 

Quest.  1 5.  Having  a  rectangular  marble  slab,  58  inches  by 
27,  I  would  have  a  square  foot  cut  off  parallel  to  the  shorter 
edge  ;  I' would  then  have  the  like  quantity  divided  from  the 
remainder  parallel  to  the  longer  side  ;  and  this  alternately 
repeated,  till  there  shall    not  be    the    quantity    of   a  foot 

left.' 


(QUESTIONS  IN  MENSURATiaN.     •  577 

left  ;  what  will  be  the  dimensions^of  the  remaining  piece  ? 

'  Ads.  20-7  inches  by  6-086. 

Q,UEST.  16.  Given  two  sides  of  an  ohtfise-an^led  triangle, 
which  are  20  and  40  poles  ;  required  the  third  side,  that  the 
Irianglg  may  contain  just  an  acre  of  land  ? 

Ans.  68-876  or  23-090. 

QvEST.  17.  The  end  wall  of  a  house  is  24  feet  6  inches 
in  breadth,  and  40  feet  to  the  eaves  ;  J  of  whicli  is  '^  bricks 
thick,  i  more  is  1^  brick  thick,  and  the  rest  1  brick  thi(  k. 
Now  the  triangular  gable  rises  38  courses  of  bricks,  4  of 
which  usually  make  a  foot  in  depth,  and  this  is  but  4^  inches, 
or  half  a  brick  thick  :  what  will  this  piece  of  work  come  to 
at  bl,  10s.  per  statute  rod  ?  Ans.  20/.  1  Is.  l^d. 

Quest.  10.  How  many  bricks  will  it  take  to  build  a  wall, 
10  feet  high,  and  500  feet  long,  of  a  brick  and  half  thick  : 
reckoning  the  brick  iO  inches  long,  and  4  courses  to  the  foot 
in  height  ?  Ans.  72000. 

Quest.  19.  How  many  bricks  will  build  a  square  pyra- 
mid of  100  feet  on  each  side  at  the  base,  and  als6  100  feet 
perpendicular  height  :  the  dimensions  of  a  brick  being  sup- 
posed 10  inches  long,  5  inches  broad,  and  3  inches  thick  ? 

An».  3840000. 

Quest.  20.  If,  from  a  right-angled  triangle,  whose  base 
is  12,  and  perpendicular  16  feet,  a  line  be  drawn  parallel  to 
the  perpendicular,  cutting  off  a  triangle  wU^e  area  is  24 
square  feet  ;  required  the  sides  of  this  trian,^le  ? 

Ans.  6,  8,  and  10. 

Quest.  21.  The  elhpse  in  Grosvenor  square  measures 
840  links  across  the  longest  way,  and  612  the  shortest,  within 
the  rails  :  now  the  walls  being  14  inches  thick,  what  ground 
do  they  enclose,  and  what  do  they  stand  upon  ? 

.         ^  enclose  4  ac  0  r  6  p. 
I  stand  on  1760|  sq.  feet. 

Quest.  22.  If  a  round  pillar,  7  inches  over,  have  4  feet 
of  stone  in  it  :  of  what  diameter  is  the  cofumn,  of  equal 
length,  that  contains  10  times  as  much  ? 

Ans.  22-136  inches. 

Quest.  23.  A  circular  fish-pond  is  to  be  made  in  a  gar- 
den, that  shall  take  up  just  half  an  acre  ;  what  must  be  the 
length  of  the  cherd  that  strikes  the  circle  ?       Ans.  27f  yards. 

V»i..  I.  74  Quest. 


&n  QUESTIONS  IN  MElNSURATION. 

Quest.  24.  When  a  roof  is  of  a  true  pitch,  or  makiog  a 
right  angle  at  the  ridge,  the  rafters  are  nearly  f  of  the 
breadth  of  the  building  :  now  supposing  the  eves-boards  to 
project  10  inches  9n  a  side,  what  will  the  new  ripping  a 
house  cost,  that  measures  32  feet  9  inches  long,  by  22  feet 
9  inches  broad  on  the  flat,  at  15s.  per  square  ? 

Ans.  SI.  15s.  9id. 

Quest.  25.  A  cable,  which  is  3  feet  long,  and  9  inches  in 
compass,  weighs  221b  ;  what  will  a  fathom  of  that  cable 
weigh,  which  measures  a  foot  about  ?  Ans.  78|  lb. 

Quest  26.  My  plumber  has  put  281b.  per  square  foot 
into  a  cistern,  74  inches  and  twice  the  thickness  of  the  lead 
long,  26  inches  broad,  and  40  deep  :  he  has  also  put  three 
stays  across  it  within,  of  the  same  strength,  and  16  inches 
deep,  and  reckons  22s.  per  cwt.  for  work  and  materials.  I, 
being  a  mason,  have  paved  him  a  workshop,  22  feet  10  inches 
broad,  with  Purbeck  stone,  at  7d.  per  foot  ;  and  on  the 
balance  I  find  there  is  3s.  6d.  due  to  him  ;  what  was  the 
length  of  the  workshop,  supposing  sheet  lead  of  j\  of  aa 
inch  thick  to  weigh  5'8991b.  the  square  foot  ? 

Ans.  32  feet,  Of  inch. 

Quest.  27.  The  distance  of  the  centres  of  two  circles^ 
whose  diameters  are  each  50,  being  given,  equal  to  30  ;  what 
is  the  area  of  the  space  enclosed  by  their  circumferences  ? 

Ans.  659- 11 9. 

Quest.  28.  If  20  feet  of  iron  railing  weigh  half  a  ton, 
when  the  bars  are  an  inch  and  quarter  square  ;  what  will 
50  feet  come  to  at  2^d.  per  lb.  the  bars  being  |  of  an  inch 
square  ?  Ans.  201.  Os.  2d. 

Quest.  29.  The  area  of  an  equilateral  triangle,  whose 
base  falls  on  the  diameter,  and  its  vertex  in  the  middle  of 
the  arc  of  a  semicircle,  is  equal  to  100  :  what  is  the  diameter 
of  the  semicircle  ?  Ans.  26-32148. 

Quest.  30.     It  is   required  to  find  the  thickness  of  the 
lead  in   a  pipe,  of  an  inch  and  quarter  bore,  which  weighs 
141b.  per  yard  in  length  j    the  eubic  foot  of  lead  weighing- 
il325  ounces  ?  Ans.  -20737  inches. 

Quest.  31.  Supposing  the  expence  of  paving  a  semicic- 
cular  plot,  at  2s.  4d.  per  foot,  come  to  10/.  ;  what  is  the 
iiameter  «f  it  ?  Ans.  14-7737  feef. 

QvissT. 


viUESTIONS  IN  MENSURATION.  679 

HuEST.  32.  What  is  the  length  of  a  chord  which  cuts  off 
^  of  the  area  from  a  circle  whose  diameter  is  289  ?      ' 

Ans.  278-6716. 

Quest.  33.  My  plumher  has  set  me  up  a  cistern,  and  his 
shop-book  being  burnt,  he  has  no  nieans  of  bringing  in  the 
charge,  and  I  do  not  choose  to  take  it  down  to  have  it  weigh- 
ed ;  but  by  measure  he  finds  it  contains  64y3_  square  feet  and 
that  it  is  precisely  |  of  an  inch  in  thickness.  Lead  was  then 
wrought  at  21/.  per  fother  of  19^  cwt.  It  is  required  from 
these  items  to  make  out  the  bill,  allowing  6f  oz  for  the  weight 
of  a  cubic  inch  of  lead  ?  Ans.  4L  1  Is.  2d. 

Quest.  34.  What  will  the  diameter  of  a  globe  be,  when  the 
solidity  and  superficial  content  are  expressed  by  the  same 
number  ?  Ans.  6. 

Quest.  35.  A  sack,  that  would  hold  3  bushels  of  corn,  is 
22i  inches  broad  when  empty  ;  what  will  another  sack  con- 
tain, which,  being  of  the  same  length,  has  twice  its  breadth, 
or  circumference  ?  Ans.  12  bushels. 

Quest.  36.  A  carpenter  is  to  put  an  oaken  curb  to  a 
round  well,  at  Sd  per  foot  square  :  the  breadth  of  the  curb 
is  to  be  71  inches,  and  the  diameter  within  3^  feet;*  what 
will  be  the  expense  ?  Ans.  bs.  2  jc?« 

Quest.  37.  A  gentleman  has  a  garden  100  feet  long,  and 
80  feet  broad  ;  and  a  gravel  walk  is  to  be  made  of  an  equal 
width  half  round  it  ;  what  must  the  breadth  of  the  walk  be 
to  take  up  just  half  the  ground  ?  Ans.  25-968  feet. 

Quest.  38.  The  top  of  a  may-pole,  being  broken  aff  by  a 
blast  of  wind,  struck  the  ground  at  10  feet  distance  from  the 
foot  of  the  pole  ;  what  was  the  height  of  the  whole  may-pole, 
supposing  the  length  of  the  broken  piece  to  be  26  feet  ? 

Ans.  50  feet. 

Quest.  39.  Seven  men  bought  a  grinding  stone,  of  60 
inches  diameter,  each  paying  }  part  of  the  expense  ;  what 
part  of  the  diameter  must  each  grind  down  for  his  share  ? 

Ans.  the  1st  4-4508,  2d  4-8400,    3d  5-3535,  4th  60765, 
5th  7-2079,  6th  9-3935,    7th  22-6778  inches. 

Quest.  40.  A  maltster  has  a  kiln,  that  is  16  feet  6  inches 
fl^aare  :  bat  ^e  wante  to  pull  it  dtwn,  aad  build  a  nevir  one, 

tlia^ 


68©  QUESTIONS  IN  MENSURATION. 

that  may  dry  three  times  as  much  at  once  as  the  ol(i  one  ;  what 
must  be  the  length  of  its  side  ?  Ans.  28  feet,  7  inches. 

Quest  41.  How  many  3-inch  cubes  may  be  cut  out  of  a 
12-inch  cube?  Ans.  64. 

Quest.  42.  How  long  must  the  tether  of  a  horse  be,  that 
will  allow  him  to  graze,  quite  round,  just  an  acre  of  ground  ? 

Ans.  39i  yards. 

Quest  43.  What  will  the  painting  of  a  conical' spire  come 
to,  at  8</.  per  yard  ;  supposing  the  height  to  be  118  feet,  and 
the  circumference  of  the  base  64  feet  ?  Ans.  141.  Os.  8|d. 

Quest.  44.  The  diameter  of  a  standard  corn  bushel  is 
18i  inches,  and  its  depth  8  ingles  ;  then  what  must  the 
diameter  of  that  bushel  be  whose  depth  is  7^  inches  ? 

Ans.  19-1067  inches. 

Quest.  45.  Suppose  the  ball  on  the  top  of  St.  Paul's 
church  is  6  feet  in  diameter  ;  what  did  the  gilding  of  it  cost 
at  3|d  per  square  inch  ?  Ans.  237^  10s.  \d. 

Quest.  46.  What  will  a  frustum  of  a  marble  cone  come 
to,  at  125.  per  solid  foot  ;  the  diameter  of  the  g|-eater  end 
being  4  feet,  that  of  the  less  end  H  ;  and  the  length  of  the 
slant  side  8  feet  ?  "  Ans.  30/.  Is.  lO^d. 

Quest.  47.  To  divide  a  cone  into  three  equal  parts  by 
sections  parallel  to  the  base,  and  to  tind  the  altitudes  of  the 
three  parts,  the  height  of  the  whole  cone  being  20  inches  ? 

Ans.  the  upper  part  13*867. 
the  middle  part  3*605. 
the  lower  part    2*528. 

Quest.  48.  A  gentleman  has  a  bowling  green,  300  feet 
long,  and  200  feet  broad,  which  he  would  raise  1  foot  higher, 
by  means  of  the  earth  to  be  dug  out  of  a  ditch  that  goes  round 
it :  to  what  depth  must  the  ditch  be  dug,  supposing  its  breadth 
to  be  every  where  8  feet  ?  Ans.  7f  |  feet. 

Quest.  49.  How  high  above  the  earth  must  a  person  be 
y^ised,  that  he  may  see  i  of  its  surface  ? 

:.  Ans.  to  the  height  of  the  earth's  diameter. 

Quest, 


QUESTIONS  IN  MENSURATION. 


^1 


"Quest.  60.  A  cubic  foot  of  brass  is  to  be  drawn  into  wire, 
ef  -1^  of  an  inch  in  diameter  ;  what  will  the  length  of  the  wire 
he,  aJlowing  no  loss  in  the  metal  ? 

Ans.  97784-7fi7  yards,  or  65  miles  984'797  yards. 


QjUEST.  51.  Of  what  diameter  must  the  bore  of  a  cannon 
be,  which  is  cast  for  a  ball  of  24lb.  weight,  so  that  the  diame- 
ter of  the  bore  may  be  -^^  of  an  inch  more  than  that  of  the  ball  ? 

Ans.  5*647  inches. 


Quest.  62.     Supposing  the  diameter  of  an  iron  91b.  ball 
to  be  4  inches,  as  it  is  very  nearly  ;  it  is  required  to  find  the 
diameters  of  the  several  balls  weighing  1,  2,  3,  4,  6,  12,  18, 
24,32,  36,  and  421b,  and  the  caliber  of  their  guns  allowing  j\  * 
of  the  caliber,  or  -^g  of  the  ball's  diameter,  for  windage. 


Answer, 


Wt.  of 

Diameter 

Caliber  of 

ball. 

ball. 

gun. 

1 

1-9230 

1-9622 

2 

2-4228 

2-4723 

3 

2-7734 

2-8301 

4 

3-0526 

3-1149 

6 

3-4943 

3-5656 

9 

4-0000 

4-oai6 

12 

4-4026 

4-4924 

18 

6-0397 

5-1425 

24 

6-5469 

5-6601 

32 

6-1051 

6-2297 

36 

6  3496 

6-4792 

42 

6-6844 

6  8208 

Quest.  53.  Supposing  the  windage  of  all  mortars  to  be 
^  of  the  caliber,  and  the  diameter  of  the  hollow  part  of  the 
shell  to  be  j\  of  the  caliber  of  the  mortar  :  it  is  required  to 
determine  the  diameter  and  weitcht  of  the  sh^ll,  and  the  quan- 
tity or  weight  of  powder  reqiiisite  to  fill  it,  foreach  ol  the  sev- 
eral sorts  ef  mortars,  namely,  the  13, 1^,  8^  6*8,  and  4-6  inch 
mortar. 

Answct; 


582 


QUESTIONS  IN  MENSURATION. 


Answer, 


Ckl'ib.  of 

Diameter 

Wt.  of  shell 

Wt.  of 

Wt    of  shell 

mort 

of  shell. 

empty. 

powder 

filled. 

4-6 

4-623 

8  320 

0-583 

8-903 

68 

5-703 

16  677 

1  168 

17-845 

43  764 

3-065 

46  829 

10 

.    9-833 

85-476 

6-986 

91  462 

13 

12783 

187  791 

13-151 

200  942 

Quest.  54.  If  a  heavy  sphere,  whose  diameter  is  4  inches, 
be  let  fall  into  a  conical  glass,  full  of  water,  whose  diameter 
is '5,  and  altitude  6  inches  ;  it  is  required  to  determine  how 
much  water  will  run  over  ? 

Ans.  26  272  cubic  inches,  or  neiarly  ^  of  a  pint^ 

Quest.  55.  The  dimensions  of  the  sphere  and  cone  being 
the  same  as  in  the  last  question,  and  the  cone  only  i  full  of 
water  ;  required  what  part  of  the  axis  of  the  sphere  is  im- 
mersed in  the  water  ?  Ans.  -546  parts  of  an  inch. 

Quest.  56  The  cone  being  still  the  same,  and  i  full  of 
water  ;  required  the  diameter  of  a  sphere  which  shall  be 
just  all  covered  by  the  water  ?  Ans.  2-445996  inches. 

Quest.  57.  If  a  person,  with  an  air  balloon,  ascend  verti- 
cally from  London,  to  such  a  height  that  he  can  just  see  Oxford 
appear  in  the  horizon  ;  it  is  required  to  determine  his  height 
above  the  earth,  supposing  its  circumference  to  be  25000 
miles,  and  the  distance  between  London  and  Oxford  49*5^38 
miles  ?  Ans.  yVoV  of  a  mile,  or  547  yards  I  foot. 

Quest.  68  In  a  garrison  there  are  three  remarkable  objects 
A,  B,  c,  the  distances  of  which  from  one  to  another  are  known 
to  be,  AB  213,  AC  424,  and  bc  262  yards  ;  1  am  desirous  of 
knowing  my  position  and  distance  at  a  place  or  station  s,  from 
which  I  observed  the  angle  asb  13^  30',  and  the  angle  csb  29^ 
50'  both  by  geometry  and  trigonometry. 
Answer, 

AS  605-7122  ;  A    C 

BS  429-6814;  ,       •  V'-^Bx 

■cs  524-2365. 

Quest.  69.  Required  the  same  as  in  the  last  question, 
when  the  point  b  is  on  the  other  side  of  ac,  supposing  ab  9, 


qjJE9TION9  IN  MENSURATION.  503 

AC  IS,  and  bc  6  furlongs  ;  also  the  angle  asb  33*  45',  and  the 
angle  bsc  22®  30'.  ^JR 

Answer, 
AS  10*64,  BS  15-64,  cs  1401. 


QjUEST.  60.  It  is  required  to  determine  the  magnitude  of 
a  cube  of  gold,  or  the  standard  fineness,  which  shall  be  equal 
to  a  sum  of  480  millions  of  pounds  sterling,  supposing  a  guinea 
to  weigh  5  dwts  9|  grains.  Ans.  18-691  feet. 

^  Quest.  61.  The  ditch  of  a  fortification  is  1000  feet  long^ 
9  feet  deep,  20  feet  broad  at  bottom,  and  22  at  top  ;  how  much 
water  will  fill  the  ditch  ? 

Ans.  1158127  gallons  nearly. 

Quest.  62.  If  the  diameter  of  the  earth  be  7930  miles, 
and  that  of  the  moon  2160  miles  :  required  the  ratio  of  their 
surfaces,  and  also  of  their  solidities  :  supposing  them  both  to 
l^e  globular,  as  they  are  ?ery  nearly  ? 

Ans.  the  surfaces  are  as  13i  to  1  nearly  : 
anil  the  soUdities  as  49^  to  1  nearly. 


is^m  Qi^  Tm  I'lRsr  volvmb. 


:$'■ 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


OCTl  6  1953  Lu 


SEP 


tlttST 


LD  21-100m-9,'48(B399sl6)476 


■y^^ft"^        f-i^it^-^f    J2      ■^'**-^yZ 


A  coursj  of  mathematics        IBIQ 


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THE  UNIVERSITY  OF  CALIFORNIA  UBRARY 


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